Demonstrate understanding of bonding, structure, properties and energy changes.
01
Before You Start
Each Chemistry standard in Level 2 is marked according to the following guideline:
Achievement is an ability to recognise facts, and define them, in isolation.
Merit is an ability to link ideas together, rather than viewing them in isolation.
Excellence is an ability to integrate different links, building a coherent answer
that responds to an unfamiliar situation.
There are also general expectations that marking reports routinely mention, which are
important to keep in mind when answering questions. These include:
Always show your working.
Units must be included for a numerical answer to be correct.
Make answers specific to the substance being discussed.
Avoid early rounding, and leave final answers to 3 significant figures.
02
Standard Contents
Like most externals, this standard will have three questions. Typically each question
will focus on one of three different topics, with these topics serving as the building
blocks of what you need to know. Hence, you need to be familiar with:
Lewis structures, polarity, and molecular shape.
Types of chemical bonding (ionic, molecular, metallic, extended covalent), and their resulting properties.
Calculations, including enthalpy change, mole calculations, and heat transfer.
03
Lessons
/Lewis Structures/
01/05
0191164 / LESSON 01
Lewis Structures
Learn how to count valence electrons, build Lewis structures,
and check that your final structure makes sense.
NCEA Level 2 Chemistry4 parts
CONTENTS
What is a Lewis structure?
91164 / LEWIS STRUCTURES / PART 01
Showing the electrons that matter
A Lewis structure is a diagram that shows how the
valence electrons of atoms are arranged in a molecule.
It shows two things:
Bonds between atoms.
Lone pairs — valence electrons that are not involved
in bonding.
KEY IDEA
Only the outer electrons matter
Lewis structures focus on valence electrons because
these are the electrons involved in chemical bonding.
Valence electrons are those in the outermost shell of an atom.
The octet rule
Most atoms are most stable when they have
8 electrons around them. This is called the
octet rule.
When drawing a Lewis structure, we therefore try to arrange
the electrons so that the atoms have a full outer shell.
Like anything in chemistry, however, there are some exceptions:
Hydrogen is the most important - it only needs 2 electrons.
Berillium - does not need a full octet, is stable with 4 or 6 electrons.
FIG. 01
How to read a Lewis structure
The diagram shows three different things:
Atoms — the elements in the molecule.
Bonds — shared pairs of electrons.
Lone pairs — non-bonding valence electrons.
A single line represents one bonding pair:
2 electrons.
Valence electrons
91164 / LEWIS STRUCTURES / PART 02
Step 1: Count the electrons
Before drawing anything, we need to know how many
valence electrons are available.
The periodic table tells us this from an element's
group number.
FIG. 01
Finding valence electrons
The periodic table is organised into
groups (columns) and periods (rows).
For the main-group elements, the group tells us
how many valence electrons the atom has.
QUICK REFERENCE
Valence electrons
Group 1 → 1
Group 2 → 2
Group 13 → 3
Group 14 → 4
Group 15 → 5
Group 16 → 6
Group 17 → 7
Group 18 → 8
For Groups 13–18, subtract 10 from the group number.
Examples
Mg is in Group 2 → 2 valence electrons.
K is in Group 1 → 1 valence electron.
N is in Group 15 → 15 − 10 =
5 valence electrons.
Cl is in Group 17 → 17 − 10 =
7 valence electrons.
IMPORTANT
For a molecule, add them together
A Lewis structure uses the total number of valence
electrons in the entire molecule.
For example, in H2O:
2 H atoms × 1 electron
+ 1 O atom × 6 electrons
= 8 valence electrons.
This total is your electron budget.
Do not lose track of it.
PRACTICE
Count the total valence electrons
Find the total number of valence electrons available
for each atom or molecule.
Mg
Magnesium
electrons
N
Nitrogen
electrons
Cl
Chlorine
electrons
H2O
2 H + 1 O
electrons
NH3
1 N + 3 H
electrons
CO2
1 C + 2 O
electrons
How to draw a Lewis structure
91164 / LEWIS STRUCTURES / PART 03
A reliable five-step method
Use the same process every time. The most important habit is to
keep track of your total valence electrons.
STEP 01
Count the total valence electrons
Add the valence electrons from every atom in the molecule.
This gives you your total electron budget.
STEP 02
Choose the central atom
Put the other atoms around a central atom.
Hydrogen is never the central atom because it can
only form one bond.
Carbon is commonly central when it is present.
Otherwise, it will typically be the atom which is alone
that is central.
STEP 03
Connect the atoms with single bonds
Draw one single bond between the central atom and each
surrounding atom.
Each bond represents 2 electrons.
Subtract 2 electrons from your total for every bond
you draw.
STEP 04
Complete the outer atoms first
Use the remaining electrons to give the surrounding
atoms full outer shells.
Add lone pairs in groups of 2 electrons.
STEP 05
Check the central atom
If the central atom does not have a full octet and
electrons are still available, convert lone pairs on
surrounding atoms into additional bonds.
This creates double or triple bonds.
Continue until the electron count is correct and the
atoms have appropriate outer shells.
FINAL CHECK
Does your structure pass all three checks?
Have I used exactly the correct number of
valence electrons?
Does each atom have the appropriate number of
electrons around it?
Does hydrogen have only 2 electrons?
If yes, your Lewis structure is ready.
FIG. 02 · WORKED EXAMPLE
Building a Lewis structure for
CH2O
Follow the method from left to right.
The diagram below shows how the electrons
are accounted for at each stage.
01
Hydrogen is different
Hydrogen only needs
2 electrons around it,
rather than a full octet.
02
Carbon is the central atom
Carbon is connected to both hydrogen
atoms and oxygen. It provides the
central framework of the structure.
03
The double bond completes the structure
After the single bonds and lone pairs
are added, the remaining electrons form
a double bond between carbon and oxygen.
Hydrogen cannot form a double bond because
it can only have two electrons around it.
PRACTICE
Draw it yourself
91164 / LEWIS STRUCTURES / PRACTICE
Your turn
For each molecule, use the five-step method.
Count the total valence electrons.
Choose the central atom.
Draw the single bonds.
Add lone pairs.
Check the electron count and octets.
Do not try to memorise the final picture.
Work it out from the electrons.
PRACTICE 01 · STARTER
Draw the Lewis structure for HCl
Hint: Hydrogen can only form one bond.
PRACTICE 02 · DEVELOPING
Draw the Lewis structure for
OCl2
Hint: Oxygen is the central atom.
Remember to give the outer atoms their full shells.
PRACTICE 03 · CHALLENGE
Draw the Lewis structure for
CO2
Hint: Start with single bonds, then check the
central carbon. You may need a multiple bond.
SECTION COMPLETE
Lewis Structures
You can now build a Lewis structure by
accounting for valence electrons, bonds,
and lone pairs.
01
Count valence electrons
Determine the total number of
valence electrons available.
02
Build the structure
Arrange the atoms and add bonds,
then distribute the remaining electrons.
03
Check the outer shells
Make sure the atoms have the
appropriate number of electrons around them.
NEXT SECTION
Molecular Shapes
→
0291164 / Lesson 02
Molecular Shapes
Use Lewis structures to determine electron geometry,
molecular shape and bond angle — then explain why
the structure has that arrangement.
NCEA Level 2 Chemistry4 parts
CONTENTS
The basic idea
THE CORE IDEA
Molecules are three-dimensional.
The atoms in a molecule are not simply arranged on a flat
page. They occupy positions in 3D space, and the
arrangement affects the molecule's properties.
Two ideas describe this arrangement:
01
Bond angle
The angle between two bonds around a central atom.
02
Molecular shape
The overall arrangement of the bonded atoms
around the central atom.
FIG. 01
Same shape, different angle
Two molecules can both have a bent molecular
shape while having different bond angles.
So, naming the shape does not always tell us the
exact angle. We need to understand what controls
the angle.
FIG. 02
Different molecular shapes
Molecular shapes can also be visibly different.
For example, a linear arrangement is different from
a bent arrangement.
The important question is:
what makes the atoms settle into these different
arrangements?
THE BIG IDEA
Electron regions repel each other.
Around a central atom, regions of electron density repel
one another. They arrange themselves so they are as far
apart as possible, giving the minimum possible repulsion.
Electron regions→Repulsion→Maximum separation→Predictable geometry
Electron geometry
STEP 01
Count the regions of electron density.
Before finding a shape or angle, count the regions of
electron density around the central atom.
A region can be:
a bonding region between the central atom
and another atom; or
a non-bonding region, also called a lone pair.
IMPORTANT
A single, double or triple bond each counts as
one region of electron density.
FIG. 01
Counting the regions
In CH2O, the double bond counts as one
region and each C–H bond counts as one region.
That gives three regions.
In H2O, the oxygen has two bonding regions
and two lone pairs. That gives four regions.
STEP 02
Use the number of regions to find the geometry.
The regions spread out to minimise repulsion. This produces
predictable arrangements called electron geometries.
RegionsElectron geometryIdeal angle
2Linear180°
3Trigonal planar120°
4Tetrahedral109.5°
WORKED EXAMPLE
NH3
Nitrogen has three N–H bonding regions and one
lone pair. That gives four regions of electron
density.
Four regions give a tetrahedral electron geometry
with an ideal angle of 109.5°.
CHECK YOURSELF
What should you be able to do now?
Count bonding and non-bonding regions.
Ignore the difference between single, double and triple bonds when counting regions.
Match 2, 3 or 4 regions to the correct electron geometry.
State the corresponding ideal bond angle.
Finding the shape
STEP 03
Electron geometry is not always molecular shape.
Electron geometry describes the arrangement of all
electron regions around the central atom.
Molecular shape describes the arrangement of the
bonded atoms only. Lone pairs are not visible as
atoms, but they still occupy space and repel the bonding
regions.
A
All regions are bonding
The molecular shape matches the electron geometry.
B
Lone pairs are present
The electron geometry stays the same, but the
observable molecular shape changes.
METHOD
Finding molecular shape from a Lewis structure
Identify the central atom.
Count all regions of electron density around
that atom.
Identify how many regions are
bonding and how many are
non-bonding.
Use the total number of regions to find the
electron geometry.
Ignore the lone pairs when naming the
molecular shape, but remember that their
repulsion affects the arrangement.
Lewis structure→Count regions→Electron geometry→Bonding vs lone pairs→Molecular shape
WHY THIS MATTERS
Trigonal planar vs trigonal pyramidal
Three bonding regions and no lone pairs give a
trigonal planar molecular shape.
Three bonding regions and one lone pair give a
trigonal pyramidal molecular shape.
Both have four or fewer electron regions involved
in the arrangement, but the lone pair changes the
observable shape.
GUIDED PRACTICE 01
H2S
Do the Lewis structure first. Then use the number of
bonding and non-bonding regions to determine the molecular
shape and bond angle.
PART A · DRAW IT
Draw the Lewis structure for H2S
PART B · SHAPE AND ANGLE
Select the molecular shape and bond angle of H2S
Molecular shape
Bond angle
GUIDED PRACTICE 02
PH3
This time, use the same method without looking at the
worked explanation first.
PART A · DRAW IT
Draw the Lewis structure for PH3
PART B · SHAPE AND ANGLE
Select the molecular shape and bond angle of PH3
Molecular shape
Bond angle
EXAM PRACTICE
Written question
THE NCEA METHOD
Explain the structure, don't just name it.
In a written question, the strongest answers connect the
Lewis structure to electron repulsion, then connect that
arrangement to the molecular shape and bond angle.
Lewis structure→Electron regions→Repulsion→Geometry→Shape + angle
State the number of regions of electron density
around the central atom.
Explain that the regions repel and arrange with
maximum separation to minimise repulsion.
Identify the electron geometry and ideal bond angle.
State how many regions are bonding and how many
are non-bonding.
Use this to identify the molecular shape.
Be specific about which molecule and central atom
you are describing.
Marking criteria, keyword check and exemplar answer
The standard
91164:
Demonstrate understanding of bonding, structure,
properties and energy changes.
This question assesses
Lewis structures and molecular shape.
Marking criteria
Achieved
Identifies the number of electron regions,
or identifies bonding and non-bonding
regions for one molecule.
Merit
Links the regions of electron density
minimising repulsion to the bond angle
for one molecule.
Excellence
Justifies the bond angle and molecular
shape of both molecules by linking
electron regions, repulsion, and
bonding versus non-bonding regions.
Keyword check
Key ideas found in your answer
Core idea
Uses regions of electron density
/ electron clouds / negative charge
Regions maximise separation
To minimise repulsion
CH2O
Identifies CH2O
Three regions of electron density
Trigonal planar geometry / shape
120° bond angle
All regions are bonding
NF3
Identifies NF3
Four regions of electron density
Tetrahedral electron geometry
109.5° bond angle
Three bonding regions
One non-bonding region / lone pair
Trigonal pyramidal molecular shape
This is a rough gauge only. It looks for particular
words and numbers, so it cannot determine whether
your reasoning is correct or which grade your answer
would earn. Compare your answer with the criteria
and exemplar.
Exemplar answer
CH2O has three regions of electron
density around the central carbon atom. The
regions repel each other and arrange themselves
with maximum separation to minimise repulsion.
This gives a trigonal planar electron geometry
with a 120° bond angle. All three regions are
bonding, so the molecular shape is also
trigonal planar.
NF3 has four regions of electron
density around the central nitrogen atom.
These regions arrange themselves to maximise
separation and minimise repulsion, giving a
tetrahedral electron geometry with an ideal
angle of 109.5°. Three regions are bonding
and one is non-bonding, so the molecular
shape is trigonal pyramidal.
SECTION COMPLETE
Molecular Shapes
You can now use electron regions and
electron-pair repulsion to determine the
shape and bond angle of a molecule.
01
Count electron regions
Identify the bonding and non-bonding
electron regions around the central atom.
02
Find the electron geometry
Use the number of electron regions to
determine how they arrange themselves.
03
Determine the molecular shape
Account for lone pairs, then identify
the molecular shape and approximate bond angle.
NEXT SECTION
Polarity
→
0391164 / Lesson 03
Polarity
Explain why bonds and molecules are polar or non-polar
using electronegativity, molecular shape and symmetry.
NCEA Level 2 Chemistry4 parts
CONTENTS
Electronegativity
THE CORE IDEA
Which atom pulls harder?
Electronegativity is the ability of an atom to attract
the shared electrons in a covalent bond.
When two different atoms form a covalent bond, they do not
necessarily attract the shared electrons equally. The atom
with the greater electronegativity pulls the electrons
closer to itself.
REMEMBER
Greater electronegativity → stronger attraction for electrons.
Why does electronegativity change?
Electronegativity depends mainly on how strongly the nucleus
can attract the outer electrons.
01
Nuclear charge
A greater positive charge in the nucleus gives the
nucleus a stronger attraction for electrons.
02
Atomic radius
If the outer electrons are further from the nucleus,
the attraction between the nucleus and those electrons
is weaker.
FIG. 01
Using the periodic table
We can use the position of elements in the periodic
table to compare their relative electronegativities.
Across a period, electronegativity generally
increases.
Down a group, electronegativity generally
decreases.
So, when comparing two elements, think about their
positions on the periodic table before anything else.
METHOD
Finding the more electronegative atom
Find both elements on the periodic table.
Compare their positions.
The element further up and to the right
generally has the greater electronegativity.
PRACTICE
Find the more electronegative element
Use the periodic table and enter the symbol of the
more electronegative element.
Sodium (Na) or Chlorine (Cl)
Fluorine (F) or Bromine (Br)
Magnesium (Mg) or Aluminium (Al)
Caesium (Cs) or Lithium (Li)
Sulfur (S) or Oxygen (O)
Carbon (C) or Nitrogen (N)
Polar Bonds
THE CORE IDEA
What happens when one atom pulls harder?
A covalent bond involves two atoms sharing electrons.
If the atoms have different electronegativities, the
electrons are shared unevenly.
The more electronegative atom attracts the shared electrons
more strongly. This gives the bond an uneven distribution
of electron density called a dipole.
01Different electronegativities
→
02Uneven electron sharing
→
03Bond dipole
Showing the dipole
Because the electrons spend more time closer to the more
electronegative atom, the two atoms develop partial charges.
δ+
Partial positive
The less electronegative atom has less electron density.
δ−
Partial negative
The more electronegative atom has greater electron density.
You may show a bond dipole using either
δ+ / δ− symbols or a dipole arrow.
WORKED EXAMPLE
H–Cl
Chlorine is more electronegative than hydrogen,
so chlorine attracts the shared electrons more strongly.
δ+
H
───────
Cl
δ−
The bond therefore has a dipole directed towards chlorine.
KEY RULE
Difference in electronegativity matters
The greater the difference in electronegativity between the
two atoms, the greater the unevenness in electron sharing
and the stronger the bond dipole.
Draw the dipole yourself
Use the tools below to annotate each bond. Show which end is
partially positive and which end is partially negative.
PRACTICE 01
H–Cl
Show the polarity of the H–Cl bond.
PRACTICE 02
N–H
Show the polarity of the N–H bond.
Polar Molecules
THE CORE IDEA
A polar bond does not automatically mean a polar molecule.
We can now identify whether individual bonds are polar.
The next question is whether the whole molecule
is polar.
To answer this, we need to consider two things:
01
Are the bonds polar?
Different electronegativities can produce bond dipoles.
02
Do the dipoles cancel?
This depends on the molecular shape and symmetry.
FIG. 01
Symmetry controls cancellation
In a symmetrical molecule, equal bond dipoles can
point in opposite directions and cancel.
In an asymmetrical molecule, the dipoles do not
cancel completely, leaving a net dipole.
Symmetrical → dipoles can cancel
Asymmetrical → dipoles do not cancel
METHOD
Finding whether a molecule is polar
Identify whether the bonds are polar.
Find the molecular shape.
Decide whether the shape is symmetrical
or asymmetrical.
Decide whether the bond dipoles cancel.
If there is a net dipole, the molecule is
polar. If the dipoles cancel, it is
non-polar.
Work through the method rather than jumping straight to
the final answer.
01
Bond polarity
O is more electronegative than H, so the O–H bonds
are polar.
02
Shape
H2O has a bent molecular shape.
03
Symmetry
The bent shape is asymmetrical, so the bond dipoles
do not cancel.
04
Conclusion
H2O has a net dipole and is therefore
polar.
Your turn: H2O
Use the Lewis structure and work through the four stages:
bond polarity, shape, symmetry, then molecular polarity.
a) Are the bonds polar?
b) Molecular shape
c) Is the shape symmetrical?
d) Is the molecule polar?
Your turn: CO2
This example is deliberately different. The C–O bonds are
polar, but the molecule can still be non-polar overall.
Work out why.
a) Are the bonds polar?
b) Molecular shape
c) Is the shape symmetrical?
d) Is the molecule polar?
EXAM PRACTICE
Written question
THE NCEA METHOD
Build the explanation, not just the answer
In a polarity question, naming the molecule as polar or
non-polar is only the starting point. Your explanation
should connect the ideas you have just learned.
PART (ii) · COMPARE AND CONTRAST THE FACTORS
THAT INFLUENCE THE POLARITY OF THESE TWO MOLECULES
0 words
Marking criteria, keyword check and exemplar answer
The standard
91164:
Demonstrate understanding of bonding,
structure, properties and energy changes.
This question assesses
Lewis structures, polarity and molecular shape.
Marking criteria
Achieved
Identifies the polarity of both molecules
(PCl3 polar,
BCl3 non-polar),
or identifies a difference in
electronegativity between the atoms
in the bonds.
Merit
Links the symmetry or asymmetry of one
molecule to the cancellation or
non-cancellation of its bond dipoles.
Excellence
Compares and contrasts both molecules,
linking electronegativity differences,
bond polarity, molecular shape,
symmetry and dipole cancellation.
Keyword check
Key ideas found in your answer
Names both molecules
Specific mention of PCl3
Specific mention of BCl3
Bond polarity
Uses the term electronegativity
Identifies a difference in
electronegativity between the atoms
Identifies the P–Cl and B–Cl
bonds as polar / having dipoles
PCl3
Trigonal pyramid shape
Asymmetrical shape
Dipoles do not cancel, leaving
a net dipole
BCl3
Trigonal planar shape
Symmetrical shape
Dipoles cancel out
The molecule is non-polar
Compare and contrast
Directly contrasts the two molecules
This is a rough gauge only. It looks for
particular words and ideas, so it cannot
determine whether your reasoning is correct
or which grade your answer would earn.
Always compare your answer with the criteria
and exemplar.
Exemplar answer
(i) PCl3 is polar.
BCl3 is non-polar.
Both the P–Cl and B–Cl bonds are polar
because there is a difference in
electronegativity between the atoms.
However, PCl3 has a trigonal
pyramidal shape, which is asymmetrical.
Therefore its bond dipoles do not cancel,
leaving a net dipole, so the molecule
is polar.
In contrast, BCl3 is trigonal
planar and symmetrical. Its bond dipoles
cancel out, so there is no net dipole and
the molecule is non-polar.
SECTION COMPLETE
Polarity
You can now use electronegativity, bond
polarity, and molecular shape to determine
whether a molecule is polar or non-polar.
01
Identify electronegativity differences
Determine whether the atoms in a bond
attract electrons equally or unequally.
02
Determine bond polarity
Identify partial charges and recognise
the direction of the bond dipole.
03
Determine molecular polarity
Use the molecular shape to decide whether
the bond dipoles cancel or leave a net dipole.
NEXT SECTION
Heat Calculations & Diagrams
→
04
91164 / Lesson 04
Heat Calculations
and Diagrams
Understand energy changes, read energy level diagrams,
and use mole, heat and bond-energy relationships to
solve NCEA calculation questions.
NCEA Level 2 Chemistry6 parts
CONTENTS
Exothermic and Endothermic
THE CORE IDEA
Where does the energy go?
A reaction or process can either take in energy from
its surroundings or release energy to its surroundings.
This gives us two important categories:
ENDOTHERMIC
Energy goes in
Energy is taken in from the surroundings.
surroundings → reaction
EXOTHERMIC
Energy goes out
Energy is released to the surroundings.
reaction → surroundings
THREE WAYS TO IDENTIFY IT
Look for the evidence
01
Temperature
If the temperature of the surroundings
increases, energy has been released:
the process is exothermic.
If the temperature decreases,
energy has been taken in:
the process is endothermic.
02
Enthalpy change
A positive ΔrH means
energy is taken in, so the reaction is
endothermic.
A negative ΔrH means
energy is released, so the reaction is
exothermic.
The overall energy change depends on
which effect is greater.
PRACTICE 01
Water that is initially frozen is heated
until it boils into a vapour.
Classify this process
GIVE A REASON
0 words
Answer
Exemplar answer
Endothermic. The water
changes state from solid to liquid
to gas. Energy must be taken in
from the surroundings to overcome
the attractions between the
particles. Since energy is being
taken in, the process is
endothermic.
PRACTICE 02
Nitrogen monoxide forms when nitrogen and
oxygen react at the very high temperatures
inside a car engine.
N2(g) + O2(g)
→ 2NO(g)
ΔrH = +180
kJ mol−1
Classify this reaction
GIVE A REASON
0 words
Answer
Exemplar answer
Endothermic.
ΔrH is positive
(+180 kJ mol−1), which
means energy is being taken in from
the surroundings. Since energy is
being taken in, the reaction is
endothermic.
PRACTICE 03
50 mL of hydrochloric acid, HCl(aq), is
added to 50 mL of sodium hydroxide, NaOH(aq),
in a polystyrene cup. The temperature of the
mixture rises from 21.0 °C to 27.5 °C.
Classify this reaction
GIVE A REASON
0 words
Answer
Exemplar answer
Exothermic. The temperature
of the mixture increased, so energy
was released from the reaction to
the surroundings. Since heat is
being released, the reaction is
exothermic.
Energy Level Diagrams
THE CORE IDEA
Showing where the energy starts and ends
An energy level diagram shows the relative energy
stored in the reactants and products.
The most important thing to remember is:
the products tell you where the reaction ends,
and the reactants tell you where it starts.
EXOTHERMIC
Products lower than reactants
Energy has been released.
ΔrH is negative.
ENDOTHERMIC
Products higher than reactants
Energy has been absorbed.
ΔrH is positive.
FIG. 01
Exothermic energy level diagram
An exothermic reaction releases heat energy
to the surroundings, so the products have
a lower energy level than the reactants.
Reactants are drawn at the higher energy.
Products are drawn at the lower energy.
ΔrH is negative.
A curved diagram may also show the
activation energy.
FIG. 02
Endothermic energy level diagram
An endothermic reaction takes in heat energy
from the surroundings, so the products have
a higher energy level than the reactants.
Reactants are drawn at the lower energy.
Products are drawn at the higher energy.
ΔrH is positive.
A curved diagram may also show the
activation energy.
NCEA CHECK
What should actually be on your diagram?
Draw and label the reactants.
Draw and label the products.
Put the products lower for an exothermic
reaction and higher for an endothermic
reaction.
Label the energy difference as
ΔrH and give the correct sign.
Use the actual substances, not just the
words “reactants” and “products”.
MELTING WATER
Reactants: H2O(s)
Products: H2O(l)
Melting is endothermic, so the products are
at a higher energy level.
Moles and Concentration
THE CORE IDEA
Counting particles using moles
A mole is a measure of the amount of substance.
In Level 2 Chemistry, you need to be able to move
fluently between amount, mass and
molar mass.
n
=
m
M
REARRANGE WHEN NECESSARY
n = m ÷ M
m = n × M
The molar mass will normally be given in the question
or resource sheet. Always check your units before
substituting numbers.
PRACTICE 01
Carbon dioxide, CO2, is produced
when fuel burns in an engine. A sample of
exhaust gas contains 300 g of CO2.
Calculate the amount, in mol, of CO2
in the sample.
M(CO2) =
44.01 g mol−1.
Give your answer to 3 significant figures.
n(CO2) =
mol
Working
Worked answer
n = m ÷ M
n =
300 ÷ 44.01 =
6.81663… =
6.82 mol
PRACTICE 02
A chemist is making a brine solution and needs
3.40 mol of sodium chloride, NaCl.
Calculate the mass, in g, of NaCl she needs
to weigh out.
M(NaCl) =
58.44 g mol−1.
Give your answer to 3 significant figures.
m(NaCl) =
g
Working
Worked answer
m =
n × M
m =
3.40 × 58.44 =
198.696 =
199 g
CONCENTRATION
When the amount is linked to a volume
The concentration of a solution tells you how many
moles of substance are dissolved in each litre.
n
=
c
×
V
WATCH THE VOLUME
1000 mL = 1.000 L
Convert mL to L before using the equation.
PRACTICE 03
A student prepares 150 mL of a sodium hydroxide
solution, NaOH(aq), with a concentration of
0.350 mol L−1.
Calculate the amount, in mol, of NaOH in
the solution.
Give your answer to 3 significant figures.
n(NaOH) =
mol
Working
Worked answer
V =
150 ÷ 1000 =
0.150 L
n =
c × V
n =
0.350 × 0.150 =
0.0525 mol
Heat Calculations
THE KEY EQUATION
Connecting heat, moles and enthalpy
Once you know the amount of reaction and the
enthalpy change per mole of reaction, you can
calculate the total heat released or absorbed.
q
=
n
×
ΔrH
qheat energy
kJ
nmoles of reaction
mol
ΔrHenthalpy change
kJ mol⁻¹
IMPORTANT
What does the “per mole” actually mean?
The r in ΔrH stands for
reaction.
ΔrH describes the heat change for
one mole of reaction. The balanced equation
tells you what that one mole of reaction involves.
2H2(g)
+
O2(g)
→
2H2O(l)
ΔrH
=
−572 kJ mol−1
Therefore one mole of reaction means:
2 mol H2 react with 1 mol O2.
2 mol H2O are produced.
572 kJ of energy is released.
Do not confuse “mol of reaction” with
“mol of a substance”.
The coefficients in the balanced equation matter.
REARRANGE
Choose the equation that matches the unknown
Find heat:
q = n × ΔrH
Find moles:
n = q ÷ ΔrH
Find enthalpy:
ΔrH = q ÷ n
Keep track of the sign of ΔrH.
A negative value represents an exothermic reaction;
a positive value represents an endothermic reaction.
Let E(C=O) be the average bond
enthalpy of the C=O bond. Write out your
working, state whether the reaction is
exothermic or endothermic, then give your
final answer with its unit.
YOUR WORKING
E(C=O) =
Five steps. Each step unlocks when the one
before it is correct.
1 · Broken
reactant bonds
2 · Made
O–H bonds
3 · Count
C=O bonds
4 · Solve
for E(C=O)
5 · Type
of reaction
STEP 1
Energy needed to break the reactant bonds
Count each type of bond in C3H8
and 5O2, multiply by its bond
enthalpy, and add them together.
bonds broken =
Σ(number × bond enthalpy)
Bonds broken =
kJ mol−1
2 × C–C = 696,
8 × C–H = 3304,
5 × O=O = 2475
696 + 3304 + 2475 =
6475 kJ mol−1
STEP 2
Energy released forming the O–H bonds
The products contain 4H2O.
Each water molecule contains two O–H
bonds.
O–H bonds made =
kJ mol−1
8 × O–H =
8 × 463 =
3704 kJ mol−1
STEP 3
Count the C=O bonds formed
The C=O bond has the unknown enthalpy,
E(C=O). How many are formed in
3CO2?
Number of C=O bonds =
3 × 2 =
6 C=O bonds
STEP 4
Find E(C=O)
The bonds made total is
3704 + 6E(C=O), and
ΔrH is
−2056 kJ mol−1.
ΔrH
=
bonds broken
−
bonds made
E(C=O) =
kJ mol−1
−2056 =
6475 −
(3704 + 6E(C=O))
6E(C=O) =
4827
E(C=O) =
4827 ÷ 6 =
804.5 =
805 kJ mol−1
STEP 5
Exothermic or endothermic?
Use the enthalpy change given in the
question.
ΔrH is negative,
so the reaction is
exothermic.
All five steps done.
Marking criteria and exemplar answer
The standard
91164:
Demonstrate understanding of
bonding, structure, properties and
energy changes.
Exemplar answer
Bonds broken =
6475 kJ mol−1
O–H bonds made =
3704 kJ mol−1
There are six C=O bonds,
giving
6E(C=O).
−2056 =
6475 −
(3704 + 6E(C=O))
E(C=O) =
805 kJ mol−1
The reaction is
exothermic because
ΔrH is negative.
SECTION COMPLETE
Heat Calculations
and Diagrams
You can now identify energy changes, represent
them on diagrams, and connect moles, heat and
bond energies in calculation questions.
01
Identify the energy change
Temperature increase or negative
ΔrH means
exothermic. Temperature decrease or
positive ΔrH means
endothermic.
02
Read the diagram
Products are lower for exothermic
reactions and higher for endothermic
reactions. Label the actual substances.
03
Use the mole equations
Know when to use
n = m ÷ M and
n = cV. Convert mL to L.
04
Use q = nΔrH
Remember that ΔrH
is per mole of reaction, so the
balanced equation matters.
05
Count bonds carefully
Bond energy calculations use
bonds broken minus bonds made.
Count every bond.
06
Finish like an NCEA answer
Show your working, include units,
avoid early rounding and give your
final answer to the requested
significant figures.
NEXT SECTION
Types of Solids and their Properties
→
05
91164 / LESSON 05
Solids and their Properties
Relate structure, bonding and particle arrangement
to the physical properties of molecular, ionic,
metallic and extended covalent solids.
NCEA Level 2 Chemistry7 parts
CONTENTS
Molecular solids
STRUCTURE → FORCES
Discrete molecules held together by
intermolecular forces.
Molecular solids contain separate molecules.
The covalent bonds inside each molecule are strong,
but the attractions between molecules are comparatively
weak.
A molecular solid is made of
individual molecules.
The covalent bonds within each molecule are
not what is normally broken when the solid melts.
Instead, the relatively weaker attractions
between molecules are overcome.
Exam habit:
identify the particles first, then identify
the force between those particles.
PROPERTIES
01
Low melting point
Usually only intermolecular forces need to be
overcome to separate the molecules.
02
Poor conductivity
There are no mobile charged particles throughout
the solid.
03
Often volatile
Weaker attractions between molecules make
separation easier.
Weak intermolecular forces→Easier to separate molecules→Lower melting point
Ionic solids
STRUCTURE → FORCES
A lattice of oppositely charged ions.
Ionic solids contain positive and negative ions arranged
in a regular three-dimensional lattice.
The structure is held together by strong
electrostatic attractions between oppositely
charged ions.
Positive and negative ions are arranged
throughout the structure.
Each ion is attracted to ions of the opposite
charge around it.
PROPERTIES
01
High melting point
Strong electrostatic attractions require
substantial energy to overcome.
02
Brittle
Shifting layers can place like charges beside
each other, causing repulsion and fracture.
03
Conducts when molten or dissolved
The ions can move when the lattice is no longer
fixed.
Fixed ions→No mobile charge→Does not conduct as a solid
Mobile ions→Charge can move→Conducts when molten / dissolved
Metallic solids
STRUCTURE → FORCES
Positive metal ions surrounded by
delocalised electrons.
Metallic bonding is the electrostatic attraction between
a lattice of positive metal ions and a sea of
delocalised electrons.
Metal atoms→Positive ions + delocalised electrons→Non-directional attraction→Metallic solid
PARTICLES
The electrons are not tied to one atom.
The metal atoms form a lattice of positive ions.
Their outer electrons become
delocalised.
These electrons can move throughout the
structure.
PROPERTIES
01
Conducts electricity
Delocalised electrons are free to move through
the structure.
02
Malleable and ductile
Layers of ions can shift while the non-directional
metallic attraction remains.
03
Often high melting point
Strong metallic attractions require energy
to disrupt.
Delocalised electrons→Electrons can move→Electrical conductivity
Extended covalent networks
STRUCTURE → FORCES
One continuous network of strong covalent bonds.
An extended covalent network has atoms joined throughout
the solid by covalent bonds.
Instead of separate molecules, the whole structure
behaves like one enormous network.
Atoms→Continuous covalent network→Strong covalent bonds throughout
EXAMPLE: CARBON
Same element. Different structure.
Diamond and graphite are both made from carbon,
but their structures differ.
Diamond forms a rigid three-dimensional network.
Graphite forms layers with weaker attractions
between the layers and delocalised electrons
within the structure.
EXAMPLE: SILICON DIOXIDE
Silicon dioxide also forms an extended covalent network.
Key idea:
it is not simply the name of the substance that matters.
It is the strength and extent of the bonding throughout
the structure.
PROPERTIES
01
Very high melting point
Strong covalent bonds throughout the network
require a large amount of energy to overcome.
02
Usually very hard
The atoms are held in a strong continuous
covalent structure.
03
Structure matters
Different arrangements of the same atoms can
produce different physical properties.
Comparing properties
THE BIG COMPARISON
Start with the structure, then explain the property.
In an NCEA question, do not simply state that a substance
has a particular property. Explain why it has that
property.
Marking criteria, keyword check and exemplar answer
The standard
91164:
Demonstrate understanding of bonding,
structure, properties and energy changes.
This question assesses
solubility, linking particles,
structure and bonding
.
Marking criteria
Achieved
• Identifies attractions are needed
between water and the substance
(KCl) for it to be soluble.
• Identifies that chlorine is
non-polar and is therefore not
attracted to water.
Merit
Links relative strengths of attractions
of the substance to water for the
solubility of ONE of the substances.
Excellence
Justifies solubility by linking particles,
structure, and bonding for both KCl
and Cl2.
Keyword check
Key ideas found in your answer
Names both substances
Specific mention of KCl
Specific mention of Cl2
KCl: particles and structure
KCl is ionic
Names the K+ ions
Names the Cl− ions
The ions separate when it dissolves
Water and the ions: attractions
Water is polar
The negative pole of water
is attracted to the K+
ions
The positive pole of water
is attracted to the Cl−
ions
Uses “attracted” / attraction
The ions are surrounded by
water molecules
Why KCl dissolves: relative strengths
Uses “force of attraction” /
electrostatic attraction
The attraction is strong enough
to overcome / stronger than
the existing attractions
Overcomes the ionic bonds
in the lattice
Overcomes the attractions
between water molecules
Cl2: why it is not soluble
Chlorine is non-polar
Chlorine is made of molecules
Chlorine is not attracted to
the polar water molecules
The attraction is not strong
enough to overcome the existing
attractions
This is a rough gauge only. It looks for
particular words and symbols, so it can't
tell whether your reasoning is correct or
which grade your answer would earn.
In particular, it can't tell “attracted”
from “not attracted”, or “strong enough”
from “not strong enough”, so check that
yourself. Compare your answer with the
criteria and the exemplar below.
Exemplar answer
Potassium chloride is ionic and when it
dissolves in water, it separates into
its ions.
The negative poles of the water molecules
are attracted to the positive K+
ions, and the positive poles of the water
molecules are attracted to the negative
Cl− ions. This causes the ions
to be surrounded by water molecules,
and the solid dissolves.
This solid is soluble because the force
of attraction between the ions and water
is strong enough to overcome the ionic
bonds in the lattice and the force of
attraction between water molecules.
The non-polar chlorine molecules are not
able to attract the polar water molecules
with sufficient strength to overcome the
solute / solute and solute / solvent
attractions and so chlorine is only
slightly soluble in water.
SECTION COMPLETE
Solids and their Properties
The property is the result of the structure.
Start with the particles and forces, then explain
what you observe.
01
Identify the structure
Decide whether the solid is molecular,
ionic, metallic or an extended covalent
network.
02
Name the particles and forces
State what particles are present and what
attractive force holds the structure together.
03
Explain the property
Link the strength and movement of particles
to melting point, conductivity, brittleness,
malleability or solubility.
Download the original NZQA papers, or sit one here: draw and write in answer boxes, submit at the end of each question, then see your marking, keyword checks and the exemplar answers. Your work saves on this device.
