Bonding
Mr Field
Using this slide show

The slide show is here to provide structure to the lessons, but not
to limit them….go off-piste when you need to!

Slide shows should be shared with students (preferable electronic to
save paper) and they should add their own notes as they go along.

A good tip for students to improve understanding of the
calculations is to get them to highlight numbers in the question and
through the maths in different colours so they can see where
numbers are coming from and going to.

The slide show is designed for my teaching style, and contains only
the bare minimum of explanation, which I will elaborate on as I
present it. Please adapt it to your teaching style, and add any notes
that you feel necessary.
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Menu:
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Lesson 1 – Ionic Bonding
Lesson 2 – Covalent Bonding
Lesson 3 – Structures
Lesson 4 – Physical Properties
Lesson 5 – Molecular Shapes
Lesson 6 – Intermolecular Properties
Lesson 7-9 – Internal Assessment
Lesson 10 – HL – Sigma and pi bonds
Lesson 11 – HL – Hybridisation
Lesson 12 – HL – Delocalisation
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Lesson 1
Ionic Bonding
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Overview
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Copy this onto an A4 page. You should add to it as a
regular review throughout the unit.
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Assessment
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This unit will be assessed by:
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An internal assessment (24%) at the end of the unit
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A joint test along with Periodicity at the end of that unit
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We Are Here
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Lesson 1: Ionic Bonding
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Objectives:
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Reflect on prior knowledge of bonding
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Refresh knowledge and understanding of ionic bonding
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Reflecting on bonding
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Brainstorm everything you already know about bonding.
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You have one minute
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Recapping ionic bonding
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An ionic bond is:
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The electrostatic attraction between two oppositely charged ions
sodium fluoride
F-
lithium oxide
Na+
Li+
O2-
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Ionic bonds typically form between a metal and a non-metal
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Ionically bonded compounds are often referred to as salts
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Li+
Ionic Structures
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Don’t worry about this now, you will be looking at in
more in Lesson 3
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Formation of simple ions
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Positive ions (cations)
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Positive ions are formed when metals lose their outer shell electrons
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Group 1: Li  Li+ + eGroup 2: Ca  Ca2+ + 2eGroup 3: Al  Al3+ + 3eTransition metals – form multiple different ions
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Fe  Fe2+ + 2eFe  Fe3+ + 3e-
Negative ions (anions)
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Negative ions are formed when non-metals gain enough electrons to
complete their outer shells
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Group 5: N + 3e-  N3Group 6: O + 2e-  O2Group 7: F + e-  F-
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Polyatomic ions
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Many ions are made of multiple atoms with an overall negative
charge
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The negative ones are mostly acids that have lost their
hydrogens
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You need to know about:
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Sulphate, SO42Phosphate, PO43Nitrate, NO3Carbonate, CO32-
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Hydrogen carbonate, HCO3Ethanoate (acetate), CH3CO2Hydroxide, OHAmmonium, NH4+
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The formula of ionic compounds
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Ionic compounds are always neutral, so the charges must balance
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Example 1: calcium reacting with fluorine:
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Calcium forms Ca2+, fluorine forms FThe formula is CaF2 so two F- charges cancel the one Ca2+
Example 2: iron (II) reacting with phosphate
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Iron (II) is the Fe2+ ion, phosphate is PO43The formula is Fe3(PO4)2
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The 6+ charges from iron (2+ x 3) balance the 6- charges (3- x 2) from phosphate
Look for the lowest common multiple
Ionic compounds do not form molecules so these are always
empirical formulae
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The names of ionic compounds
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The cation gives the first part of the name
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Normally a metal except in the case of ammonium (NH4+)
In the case of transition metals, Roman numerals tell you the charge
on the metal ion
The anion gives the second part of the name
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Simple ions: ‘-ide’…e.g. chloride, fluoride, nitride etc
Complex ions: just their name: sulphate, phosphate etc
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Note: the ‘-ate’ ending usually refers to polyatomic ions containing
oxygen, which provides the negativity…more on this in the redox unit
Examples:
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CaF2: calcium fluoride
Fe3(PO4)2: iron (II) phosphate
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Your turn
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Deduce the formulae and names of the ionic compounds
formed between:
1.
Lithium and fluorine
6.
Sodium and sulphate ions
2.
Magnesium and iodine
7.
Chloride and ammonium ions
3.
Aluminium and oxygen
8.
Iron (III) and sulphate ions
4.
Iron (II) and sulphur
9.
Iron (II) and nitrate ions
5.
Calcium and nitrogen
10.
Potassium and carbonate ions
11.
Work through the simulation here:
http://www.learner.org/interactives/periodic/groups_interactive.html
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Key Points
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Ionic bonds are the attraction between two oppositely
charged ions
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Ionic bonds form between metals and non metals
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Metals lose their outer shell
Non-metals complete their outer shell
The number of each ion in the formula is determined by
the lowest common multiple of their charges
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Homework
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Research and make notes on metallic bonding. Including:
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Description of the nature of the metallic bond
Factors affecting the strength of metallic bonds
Explanation of the malleability of metals
Explanation of the electrical conductivity of metals
Factors affecting the conductivity of metals
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Lesson 2
Covalent Bonding
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Refresh
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Predict and explain which of the following compounds are
ionic:
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NaCl
BF3
CaCl2
N2O
P4O6
FeS
CBr4.
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Lesson 2: Covalent Bonding
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Objectives:
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Refresh knowledge and understanding of covalent bonding
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Learn how to draw Lewis structures
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Identify examples of dative bonding
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Identify instances of expanded octets
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Recapping ionic bonding
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An ionic bond is:
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The electrostatic attraction between two oppositely charged ions
sodium fluoride
F-
lithium oxide
Na+
Li+
O2-
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Ionic bonds typically form between a metal and a non-metal
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Ionically bonded compounds are often referred to as salts
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Li+
Covalent bonding
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A covalent bond is the attraction of two atoms to a shared pair of
electrons
water
Each O has two single bonds
H
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O
H
C
O
Atoms aim for complete outer-shells, and each covalent bonds gives
them one electron
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O
carbon dioxide
each C has two double bonds
Atoms form as many bonds as they have gaps in their outer-shells
Covalent bonds typically form between two non-metals
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How many bonds?
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Atoms (usually) form bonds according to the ‘octet’ rule
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Atoms form as many bonds as they have ‘gaps’ in their outer shells, with
each bond gaining them one electron:
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This means they try to get a full outer shell of 8 electrons (except hydrogen
which is full at 2)
Group 7: 7 electrons, 1 gap  1 bond
Group 6: 6 electrons, 2 gaps  2 bonds
Group 5: 5 electrons, 3 gaps  3 bonds
Group 0/8: 8 electrons, 0 gaps  0 bonds
Covalent bonds can be:
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Single: one shared electron pair, X-X
Double: two shared electron pairs, X=X
Triple: three shared electron pairs, XX
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Covalent Structures
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Molecular
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Giant lattice
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As in water and methane
As in silicon dioxide
More on these later in the unit
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Lewis structures
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Show the position of outer-shell electrons in a covalent
compound
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Various types: all show the same thing, any is fine
dots and crosses crosses only
dots only
lines
Blue Circles: These are the bonding pairs of electrons – the ones involved in the
bonds.
Red Circles: These are non-bonding or lone pairs of electrons. They are very
important, but students often forget about them!
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Working out a Lewis structure
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Example: diazene, N2H2
Don’t worry about
the shape…more
on that later!
Step 1: Write the number of
Nitrogen: 5 electrons, 3 bonds
electrons in each atom and the
Hydrogen: 1 electron, 1 bond
number of bonds each atom can form
Step 2: Draw the structure using
lines for bonds
There will be 2 N-H bonds and 1 N=N bond
Step 3: Add in the lone pairs
The N started with 5 electrons, and 3 are in
bonds, so that leaves 2 remaining…each N
will have one lone pair
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Time to practice…again
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Draw Lewis structures for the following, bearing in mind
the previous two slides
1.
H2
6.
NH3
2.
O2
7.
CO2
3.
N2
8.
HCN
4.
H2O
9.
C2H4
5.
HCl
10.
C2H2
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The dative-bond
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Sometimes an atom will contribute both of the electrons in a
covalent bond, this is called a dative (covalent) bond
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E.g.
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In this example, the lone pair from a water molecule has
formed a dative bond to a hydrogen ion (H+)
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You can show dative bonds with an arrow to say where the
electrons came from…but do not have to
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The expanded octet
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In this example, the Lewis structure of
SO3 shows it with 12 electrons in the
outer shell
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This is because sulphur can make use of
its empty d-orbitals (the 3d ones)
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This is called an expanded octet
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Period 2 elements can’t do this as they have
no d-orbitals
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Time to practice…again
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Draw Lewis structures for the following, bearing in mind
the previous two slides
1.
NH4+
5.
SF6
2.
SO2
6.
PCl5
3.
B2F6
7.
CO
4.
Al2Cl6 (yes covalent!)
8.
XeF4
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Homework
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Check tables 9 and 10 of the data booklet
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Draw a graph of bond length vs. bond enthalpy (strength)
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The easiest way is to enter the data into Excel and get it to
draw it
Identify and explain the relationship between bond length
and bond strength.
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Identify any significant exceptions to this trend and explain why
they occur.
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Key Points
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Atoms (generally) form covalent bonds according to the
octet rule.
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Each covalent bond gives an atom one extra electron
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In dative bonds, both the electrons in the bond come
from the same atom
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Period 3 (and above) elements can break the octet rule
by using empty d-orbitals and might have 12 or more
electrons in their outer shell
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Lesson 3
Structures
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Refresh
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How many lone pairs and bonding pairs of electrons
surround xenon in the XeF4 molecule?
Lone pairs
A.
4
B.
0
C.
0
D.
2
Bonding pairs
8
8
4
4
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Lesson 3: Structures
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Objectives:
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Describe and compare the structures and properties of:
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Allotropes of carbon
Silicon and silicon dioxide
Ionic compounds
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Marketplace – in four groups
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Each group needs to produce a learning resource to teach the other
students about their chosen topic.
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Once the resources are completed, one person should remain with the
resource whilst the remaining members circulate and learn from the other
stations….you should manage your time, taking turns manning your station
to make sure everyone makes it round class.
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There will be a test at the end.
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Groups should look at the structure, bonding, properties and uses of:
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Group 1: Diamond and graphite
Group 2: Buckminster Fullerenes and carbon nanotubes
Group 3: Silicon and silicon dioxide
Group 4: Ionic compounds (less focus on uses)
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Test Time
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Complete the test here
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You have 10 minutes
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Alternative to Marketplace Activity
1.
2.
3.
4.
Produce a table summarising the differences in properties
of three allotropes of carbon: diamond (this should include
silicon and silicon dioxide which have similar structures),
graphite and buckminster fullerene. Look at structure,
properties (explained), uses (and how these are related to
the properties)
Draw a mind-map summarising the three main types of
structure: giant ionic, giant covalent, molecular including a
drawing of each, three example compounds, and an
explanation of their properties.
Produce a Venn Diagram to summarise then similarities
and differences between the three main types of bonding
Produce a flow chart that can be followed to determine
the type of bonding present in an element/compound and
(for ionic/covalent) produce Lewis diagrams to describe it.
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Homework
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Graphene is a recently discovered ‘super-material’
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Research its structure and properties and as many
potential uses for it as possible
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Make sure you have
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Key Points
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Carbon (diamond, graphite and fullerenes), silicon and
silicon dioxide exhibit giant covalent (macromolecular)
structures
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For example diamond: each carbon is bonded to exactly four
others and so on
Ionic compounds form giant ionic lattices
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NaCl: every Na+ ion surrounded by 6 Cl- ions, every Cl- ion
surrounded by 6 Na+ ions
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Lesson 4
Physical Properties
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Refresh
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Unlike many covalently bonded substances, graphite is an
excellent conductor. Describe the structure and bonding
in graphite and explain why it is such a good conductor.
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Lesson 4: Testing Properties
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Objectives:
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Design and conduct an experiment to use the physical
properties of compounds to differentiate between them
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Design Challenge
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You will be provided with samples of 6 unknown substances:
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Lead shot, iodine granules, graphite, sodium chloride, sugar, lead
bromide
You need to design and conduct a series of experiments to let
you determine the nature of the bonding in each, using their
physical properties. Focus on:
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Solubility in polar solvents (such as water)
Solubility in non-polar solvents (such as hexane)
Melting/boiling point
Electrical conductivity (when solid, molten or in solution)
Volatility (evaporatingness…not a word but you know what I mean!)
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Homework
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Bring laptop to next lesson and make sure you have
installed ACD/Labs ChemSketch (Google it!)
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Identify which substance is which out of:
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Lead
Graphite
Iodine
Sugar
Sodium chloride
Lead bromide
Fully explain your reasoning
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Key Points
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The structure and bonding of substances has a very
significant effect on their properties
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Lesson 5
Molecular Shapes
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Refresh
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State two physical properties associated with metals and
explain them at the atomic level.
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Lesson 5: Molecular Shapes
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Objectives:
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Model, draw, and explain the shapes of molecules with 2-4
charge centres around a central atom
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HL Only: Model, draw, and explain the shapes of molecules
with 5 and 6 charge centres around a central atom
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VSEPR – a brief introduction
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Valence Shell Electron Pair Repulsion (aka ‘vesper’)
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Pairs of electrons around an atom repel each other – this determines a molecules
shape
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Pairs of electrons are known as ‘charge centres’ and include both:
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The electrons in a covalent bond
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a double/triple bond only counts as one charge centre!
Lone pairs / non-bonding pairs
Example: ethyne
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The carbon has two charge centres (the C-H bond and the CC bond)
They push as far away from each other as possible making a 180o bond angle
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Drawing shapes in three dimensions:
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Draw as many atoms as you
can in the same plane ‘flat’ on
the paper
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Use solid wedges to show
atoms coming out of the
plane of the paper towards
you
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Use dashed wedges to show
atoms going back into the
plane of the paper away from
you
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Modelling Molecules
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You need to model molecules with varying numbers of charge centres. For each
one you should:
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HL and SL: two, three and four charge centres including:
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CO2, C2H2, BF3, C2H4, SO2, CH4, NH3, H2O, NH4+, H3O+
HL only: five and six charge centres including:
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Draw a Lewis structure
Draw the structure (in 3D where relevant)
Label bond angles and explain them
Name the structure
PCl5, SF6, XeF4, PF6-
How:
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Complete the Modelling Molecular Shapes activity here
It is best to start with the ones in bold as typical examples
If you like, you could also make models by tying balloons together at their centre and
observing the shapes they form
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Key Points
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Negative charge centres repel each other, this determines the shape
of a molecule
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Standard shapes:
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Two charge centres: linear
Three charge centres: trigonal planar
Four charge centres: tetrahedral
Five charge centres: trigonal bipyramidal
Six charge centres: octahedral
Lone pairs:
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Start with the above shape and then take off one bond
Bond angles compressed as lone pairs a concentrated source of charge
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Lesson 6
Intermolecular Forces
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Refresh

Use the VSEPR theory to deduce the shape of H3O+ and
C2H4. For each species, draw the Lewis structure, name
the shape, and state the value of the bond angle(s).
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We Are Here
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Lesson 5: Molecular Shapes
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Objectives:
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Learn to identify and explain the three types of intermolecular
forces:
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Van der Waals
Permanent dipole-dipole
Hydrogen bonds
Understand and explain the effects of the above on
melting/boiling points
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Intermolecular Forces
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The attractive forces between
molecules
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It is these that are partially broken
during melting, and fully broken
during boiling
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Note: when molecular compounds
melt/boil, the bonds in the
molecule do not break, it is just the
attractive forces between the
molecules that break
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Van der Waals Forces aka
Temporary dipole induced dipole forces
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Random electron movements create a small,
temporary dipole
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This induces a similar dipole in a
neighbouring molecule
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This creates a small attraction between them
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These are weak and exist only for the tiniest
fraction of a second
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Van der Waals forces are present in all
molecules
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Van der Waals forces:
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Increase with molecular mass
Decrease with the roundness of a molecule
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Dipole-dipole forces
aka
Permanent dipole forces
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Different atoms have different electronegativities, which
means there will be variations in the electron charge
density in different parts of a molecule
-
If a molecule is not symmetrical, the variation produces
a dipole where a molecule as a positive and a negative
end
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+
The end with high charge density is The end with low charge density is +
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Oppositely charged dipoles attract each other.
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This is a relatively strong attractive force
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If a molecule is symmetrical, variations in electron
charge density cancel each other out and the molecule
is non-polar
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-
+
-
Hydrogen bonds aka
H-bonds
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The strongest type of intermolecular force

They occur between a nitrogen, oxygen or
fluorine and a hydrogen that is bonded to a
nitrogen, oxygen or fluorine

N, O and F are the three most electronegative
elements, and all have lone-pairs when bonded

When H is bonded to N, O or F, the electrons
in the bonded are strongly attracted to the
N/O/F, leaving the H very positive

The lone pair on the N/O/F is strongly
attracted to the positive hydrogen
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Effects of intermolecular forces
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Intermolecular forces play an important role in the
properties of compounds including:
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Melting/boiling point:
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Volatility:
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Stronger intermolecular forces  higher m.p./b.p.
Stronger intermolecular forces  lower volatility
Solubility: like dissolves in like
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Polar solutes dissolve best in polar solvents
Non-polar solutes dissolve best in non-polar solvents
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Looking into intermolecular forces

Complete the activity here to research and model
intermolecular forces
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Summary
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Three types of intermolecular force, from strongest to
weakest:
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Hydrogen bonds
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Dipole-dipole
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Between N/O/F and H attached to N/O/F
Between permanent dipoles on asymmetric molecules
Van der Waals

Between instantaneous dipoles formed on any molecule/atom
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Lessons 7-9
Intermolecular Forces Internal Assessment
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Refresh

Which compound has the highest boiling point?
A.
B.
C.
D.
CH3CH2CH3
CH3CH2OH
CH3OCH3
CH3CHO
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We Are Here
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Lessons 7-9: Internal Assessment

Objectives:

Understand the requirements of an internal assessment

Design and conduct an internal assessment on intermolecular
forces
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IB internal assessment

24% of your final grade will come from the internal assessment portfolio.

You are awarded marks in 5 areas out of a possible maximum of 48 marks:

Design: 2 x 6  12 marks
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Data Collection and Processing: 2 x 6  12 marks
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The two best marks from any of your assignments
Interpersonal Skills: 6 marks
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The two best marks from any of your assignments
Conclusion and Evaluation: 2 x 6  12 marks
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The two best marks from any of your assignments
Assessed by the Group 4 project
Manipulative Skills: 6 marks

Assessed cumulatively over the whole course
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An exemplar
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Read through the assessment criteria crib-sheet here
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Read through the example internal assessment here
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The topic
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The topic for your internal assessment should be:
‘a factor affecting intermolecular forces’
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Defining your research question

Things to consider:
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Your question must be tight and clearly focussed
It should be specific, (correctly) naming the chemical compounds
involved
It should explicitly suggest the independent variable
It should implicitly or explicitly suggest the dependent variable
Choice of independent variable:
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Begin with the end in mind – choose a smoothly variable variable
that can be graphed on a scatter-gram
Choose variables that are straightforward to fully control
Choose variables with at least 5 suitable variations
You could consider families of related compounds or varied mixtures
of two or more compounds
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Designing your experiment

Ensure you consider all the variables that you will control and
state how you will control them

Be suitably precise with your method:

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1.00 cm3 is precise, 1 cm3 is not!
0.140167 mol is too precise, 0.140 mol is suitably precise

State explicitly which variations you will do and how many
repetitions you will do

Conduct trial runs to help you refine the method you will
use…this takes more time initially but saves it in the long run
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Conducting your experiment

Record qualitative data (descriptions) in addition to
qualitative data

Record all quantitative data in clearly labelled tables AS
YOU DO THE EXPERIMENT


Table headings should include the uncertainty of measurements
Make sure you take note of the uncertainty of each of the
items of apparatus used to take a measurement
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Data processing

Work through and explain one example only of each of the calculations that
need performing

Produce a table showing the intermediate steps in your calculations and the final
answers

It is generally appropriate to do calculations based on the average of your raw data
rather than doing all calculations on all your data and then averaging

You must calculate the uncertainty in your final answers



Work through and explain one example of the calculation used
Present this as a final column on your table, to the right of your ‘final answer’ column
You should draw a scatter graph (nearly always) of your independent variable and
your final results

This should include an appropriate line/curve of best fit and error bars
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Conclusion and evaluation

Conclusion
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
You need to identify and explain the chemistry behind the trend in
your results
If this trend is not what you expected, you should account for
why…this may be because you got the chemistry wrong, it may be
due to flaws in the experiment
You should compare your results to those in the literature
Evaluation:


You need to comment on how effective your experiment was,
identifying weaknesses and limitations
You should identify specific and detailed improvements to the
experiment to overcome the above….be bold: saying do more
repeats or variations is not enough
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Lesson 10
HL Only
Sigma and Pi Bonds
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We Are Here
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Lesson 10: Sigma and Pi Bonds

Objectives:

Understand and identify pi and sigma bonds

Meet molecular orbital theory and attempt some simple
molecular orbital diagrams
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Sigma-bonds/ -bonds

-bonds are formed when two
orbitals from neighbouring
atoms overlap along the axis of
approach

This creates a new orbital called
a -bonding orbital


You can just call it a -bond
Viewed from the side, bonding orbitals have the
shapes shown in the red box


Viewed end on, -bonding
orbitals appear circular
Check the visualisation here:
http://www.falstad.com/qmmo/
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Pi-bonds/π-bonds


π-bonds are formed when two orbitals
from neighbouring atoms overlap
perpendicular to the axis of approach

This creates a new orbital called a πbonding orbital

You can just call it a π–bond

Viewed from the side, π-bonding orbitals have the shape shown on the right

Viewed end on, π-bonding orbitals appear as two circles, one above the atoms and
one below

π -bonds make up the second and third bonds of double and triple bonds

π -bonds are weaker than -bonds due to the weaker overlap of the orbitals
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Counting Bonds

For each of the following, determine the number of pi and
sigma bonds:
1.
H2
1.
N2
2.
CH4
2.
C2H4
3.
O2
3.
C3H8
4.
H2O
4.
C3H6
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Molecular Orbital Theory

This theory suggests that when molecules bond, atomic orbitals merge to
form molecular orbitals.

Molecular orbitals are like regular orbitals (they have specific energy levels,
can hold 2 electrons etc), but they extend over more than one atom.

Electrons have some wave-like properties (thank quantum mechanics for
this insight), thus when two orbitals overlap their waves interfere
(http://phet.colorado.edu/en/simulation/wave-interference):

Constructive interference  bonding orbital


Creates an attractive force between atoms
Destructive interference  anti-bonding orbital

Creates a repulsive force between atoms
Note: this goes beyond the syllabus
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The H2 molecule

The 1s orbitals from
each H overlap to create
two new molecular
orbitals:



bonding orbital
* anti-bonding orbital
Both 1s electrons go into
the bonding orbital,
with none in the *
anti-bonding orbital, so
there is a net attractive
force
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The He2 molecule

Similar to H2, the and *
molecular orbitals are created

This time, two electrons go into
each one

The attractive effect of the full
bonding orbital is exactly
balanced by the repulsive effect of
the full * anti-bonding orbital so
there is no overall attraction

Thus He2 does not exist!
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The O2 molecule

The main thing of interest
here is the overlap of the porbitals creating:


1 and 2 π bonding orbitals
1 * and 2 π* anti-bonding
orbitals

The -p orbital is filled, and
the two π-orbitals are filled
but the two π* orbitals are
only partially filled

The net effect is that the O
atoms are connected by one
and one π bond
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Have a go yourself

Attempt to draw molecular orbital diagrams for the
following, and determine the number of and π bonds

F2

N2

NO
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Refresh

bonds form from the overlap or orbitals along the axis

π bonds are weaker and form from the overlap of orbitals
perpendicular to the axis

Molecular orbital theory involves the constructive and
destructive interference of atomic orbitals
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Lesson 11
HL Only
Sigma and Pi Bonds
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Refresh

How many sigma (σ) and pi (π) bonds are present in the
structure of HCN?
A.
B.
C.
D.
σ
1
2
2
3
π
3
3
2
1
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Lesson 11: Hybridisation

Objectives:

Understand the formation of hybrid orbitals

Identify the hybridisation of atoms

Understand the causes and effects of hybridisation
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A challenge:

Before the development of molecular orbital
theory, it was believed that bonds formed from
the overlap of partially filled orbitals.

Draw a 3D (ish) diagram of the valence orbitals
in a carbon atom, showing the electrons in each

Now attempt to show the overlap with the sorbitals of 4 H atoms to create methane
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Hybridisation

One effect of the wave-nature of electrons is molecular
orbitals (as mentioned last lesson)

Another is hybridisation in which s and p orbitals merge
together to create new atomic orbitals

This is important as without it we can’t explain the bonding in
molecules such as methane

We need to know about three types of hybridisation:



sp3
sp2
sp
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sp3 hybridisation

One ‘s’ orbital combines with
three ‘p’ orbitals to create four
‘sp3’ orbitals

The sp3 orbitals are arranged
tetrahedrally

Most atoms with tetrahedral
geometry will be sp3 hybridised

Check the visualisation here:
http://www.uwosh.edu/faculty_s
taff/gutow/Orbitals/N/sp3%20hy
brid.shtml
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Two more hybrids

In addition to sp3, there is also sp2 and sp hybridisation

Try to draw diagrams similar to this to show how the
orbitals combine

Use the visualiser here
(http://www.uwosh.edu/faculty_staff/gutow/Orbitals/N/sp
2%20hybrid.shtml ) for each one to draw and label both
the shape of the individual orbital and the overall 3D
arrangement of orbitals around the atom
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sp2 hybridisation

One ‘s’ orbital combines with
two ‘p’ orbitals to create three
‘sp2’ orbitals, leaving one ‘p’
orbital untouched

The sp2 orbitals have a trigonal
planar arrangement

Most atoms with trigonal planar
geometry will be sp2 hybridised

Check the visualisation here:
http://www.uwosh.edu/faculty_s
taff/gutow/Orbitals/N/sp2%20hy
brid.shtml
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sp hybridisation

One ‘s’ orbital combines with
one ‘p’ orbital to create two ‘sp’
orbitals and leaving two ‘p’
orbitals untouched

The sp orbitals have a linear
arrangement

Most atoms with linear
geometry will be sp hybridised

Check the visualisation here:
http://www.uwosh.edu/faculty_s
taff/gutow/Orbitals/N/sp%20hyb
rid.shtml
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Task: Hybridisation and bonding


Draw diagrams (in 3D) showing the
hybridisation and orbital over-lap, and
where the electrons are, in the
bonding of:

CH4

C2H4

C2H2

N2

H2 O

CO2

Using the C-C bond and hybridisation from
an ethane molecule as an example, try to
work out and explain why a π-bond
couldn’t form with an sp3 hybridised carbon.
Molecular models may help.

Copy each formula, write the hybridisation
next to each carbon and draw in the bond
angles around each carbon
Label each orbital (either s, p, sp3 sp2
or sp), indicate whether each bond is
a -bond or a π-bond and label the
hybridisation around each atom
(except H).
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Key Points

Tetrahedral atoms have sp3 hybridisation

Trigonal planar atoms have sp2 hybridisation

Linear atoms have sp hybridisation
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Lesson 12
HL Only
Delocalisation
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Refresh

What is the type of hybridization of the silicon and
oxygen atoms in silicon dioxide?
A.
B.
C.
D.
Silicon
sp3
sp3
sp2
sp2
Oxygen
sp3
sp2
sp3
sp2
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Lesson 12: Delocalisation

Objectives:

Understand and identify pi and sigma bonds

Meet molecular orbital theory and attempt some simple
molecular orbital diagrams
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Benzene

Benzene, C6H6 is a cyclic compound containing a 6carbon ring

Draw a possible structure for benzene, labelling the
hybridisation of each carbon, the bond angles and
researching the bond lengths in the data booklet
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Delocalisation
(http://www.chemtube3d.com/orbitalsbenzene.htm)

The top two diagrams show the remaining carbon porbitals in benzene, which can overlap to form π-bonds

When the p-orbitals overlap to form a π-bond, they
could:



Overlap ‘left’ (like the top diagram)
Overlap ‘right’ (like the middle diagram),
However actually:

They overlap in both directions, creating a doughnut-shaped
cloud of electrons above and below the ring (bottom)

This ‘π-cloud’ contains 6 electrons in total, all free to
move

The electrons in the π-cloud are delocalised – free to
move anywhere in the cloud
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Resonance

An alternative way to think of delocalisation is resonance

In resonance, we can think of species as constantly
flipping between two (or more) equivalent forms, so that
on average they are half-way between.
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Effects of Delocalisation

When we measure the carbon-carbon bond
lengths in benzene, we find they are all equal at
140 pm

Compare this to:


C=C bond – 135 pm
C-C bond – 147 pm

Delocalisation explains this as now rather than
having 3 C=C and 3 C-C, we have 6 C-C where
each bond is the equivalent of 1 ½ bonds

Delocalisation makes species more stable by
spreading out the electron charge making it less
attractive to positive attackers (electrophiles)

It is the stability of benzene that makes it such a
serious carcinogen
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Identifying delocalisation

Delocalisation can happen anywhere you have a neighbouring
atoms that are equivalent to each other, allowing pi bonds to
form either direction.

The following species also exhibit delocalisation:


NO3-, NO2-, CO32-, O3, RCOO-
For each one:





Draw a Lewis structure of each of the possible resonance forms
Draw a Lewis structure using --- lines to show the delocalisation
Draw a three dimensional structure showing how the overlapping of
p-orbitals leads to the delocalisation of the π-electrons
State the number of electrons in the π-cloud
Research the lengths of the bonds in the delocalised and nondelocalised forms
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Extension

The delocalisation of π-electrons is responsible for the
colour of many organic compounds

Find three organic (carbon-containing) dyes online, draw
their structures and then draw in the delocalised πsystem
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Key Points

Delocalisation happens when pi-bonds extend across
more than one carbon

Delocalisation increases the stability of molecules and
ions
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