Bonding

Mr Field

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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: 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 Copy this onto an A4 page. You should add to

it as a regular review throughout the unit.

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Assessment

This unit will be assessed by:

An internal assessment (24%) at the end of the unit

A joint test along with Periodicity at the end of that unit

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Lesson 1: Ionic Bonding

Objectives:

Reflect on prior knowledge of bonding

Refresh knowledge and understanding of ionic bonding

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Reflecting on bonding

Brainstorm everything you already know about bonding.

You have one minute

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Recapping ionic bonding An ionic bond is:

The electrostatic attraction between two oppositely charged ions

sodium fluoride lithium oxide

Ionic bonds typically form between a metal and a non-metal

Ionically bonded compounds are often referred to as salts

Li+O2-Li+Na+F-

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Ionic Structures Don’t worry about this now, you will be

looking at in more in Lesson 3

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Formation of simple ions Positive ions (cations)

Positive ions are formed when metals lose their outer shell electrons Group 1: Li Li+ + e-

Group 2: Ca Ca2+ + 2e-

Group 3: Al Al3+ + 3e-

Transition metals – form multiple different ions Fe Fe2+ + 2e-

Fe Fe3+ + 3e-

Negative ions (anions) Negative ions are formed when non-metals gain enough

electrons to complete their outer shells Group 5: N + 3e- N3-

Group 6: O + 2e- O2-

Group 7: F + e- F-

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Polyatomic ions Many ions are made of multiple atoms with an overall

negative charge

The negative ones are mostly acids that have lost their hydrogens

You need to know about: Sulphate, SO4

2-

Phosphate, PO43-

Nitrate, NO3-

Carbonate, CO32-

Hydrogen carbonate, HCO3-

Ethanoate (acetate), CH3CO2-

Hydroxide, OH-

Ammonium, NH4+

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The formula of ionic compounds Ionic compounds are always neutral, so the charges must

balance

Example 1: calcium reacting with fluorine: Calcium forms Ca2+, fluorine forms F-

The formula is CaF2 so two F- charges cancel the one Ca2+

Example 2: iron (II) reacting with phosphate Iron (II) is the Fe2+ ion, phosphate is PO4

3-

The formula is Fe3(PO4)2 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 The cation gives the first part of the name

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 Simple ions: ‘-ide’…e.g. chloride, fluoride, nitride etc Complex ions: just their name: sulphate, phosphate etc

Note: the ‘-ate’ ending usually refers to polyatomic ions containing oxygen, which provides the negativity…more on this in the redox unit

Examples: CaF2: calcium fluoride Fe3(PO4)2: iron (II) phosphate

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Your turn Deduce the formulae and names of the ionic

compounds formed between:1. Lithium and fluorine

2. Magnesium and iodine

3. Aluminium and oxygen

4. Iron (II) and sulphur

5. Calcium and nitrogen

6. Sodium and sulphate ions

7. Chloride and ammonium ions

8. Iron (III) and sulphate ions

9. Iron (II) and nitrate ions

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

Ionic bonds are the attraction between two oppositely charged ions

Ionic bonds form between metals and non metals 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

Research and make notes on metallic bonding. Including: 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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Predict and explain which of the following compounds are ionic: NaCl BF3 CaCl2 N2O P4O6 FeS CBr4.

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Lesson 2: Covalent Bonding

Objectives:

Refresh knowledge and understanding of covalent bonding

Learn how to draw Lewis structures

Identify examples of dative bonding

Identify instances of expanded octets

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Recapping ionic bonding An ionic bond is:

The electrostatic attraction between two oppositely charged ions

sodium fluoride lithium oxide

Ionic bonds typically form between a metal and a non-metal

Ionically bonded compounds are often referred to as salts

Li+O2-Li+Na+F-

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Covalent bonding A covalent bond is the attraction of two atoms to a shared pair of

electrons

water carbon dioxideEach O has two single bonds each C has two double bonds

Atoms aim for complete outer-shells, and each covalent bonds gives them one electron Atoms form as many bonds as they have gaps in their outer-shells

Covalent bonds typically form between two non-metals

H HO O OC

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How many bonds? Atoms (usually) form bonds according to the ‘octet’ rule

This means they try to get a full outer shell of 8 electrons (except hydrogen which is full at 2)

Atoms form as many bonds as they have ‘gaps’ in their outer shells, with each bond gaining them one electron: 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: 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

Molecular As in water and methane

Giant lattice As in silicon dioxide

More on these later in the unit

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Lewis structures

Show the position of outer-shell electrons in a covalent compound

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 Example: diazene,

N2H2Step 1: Write the number of electrons in each atom and the number of bonds each atom can form

Nitrogen: 5 electrons, 3 bondsHydrogen: 1 electron, 1 bond

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

Don’t worry about the

shape…more on that later!

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Time to practice…again Draw Lewis structures for the following,

bearing in mind the previous two slides1. H2

2. O2

3. N2

4. H2O

5. HCl

6. NH3

7. CO2

8. HCN

9. C2H4

10. C2H2

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The dative-bond Sometimes an atom will contribute both of the

electrons in a covalent bond, this is called a dative (covalent) bond

E.g.

In this example, the lone pair from a water molecule has formed a dative bond to a hydrogen ion (H+)

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

In this example, the Lewis structure of SO3 shows it with 12 electrons in the outer shell

This is because sulphur can make use of its empty d-orbitals (the 3d ones)

This is called an expanded octet Period 2 elements can’t do this as

they have no d-orbitals

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Time to practice…again Draw Lewis structures for the following,

bearing in mind the previous two slides1. NH4

+

2. SO2

3. B2F6

4. Al2Cl6 (yes covalent!)

5. SF6

6. PCl5

7. CO

8. XeF4

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Homework

Check tables 9 and 10 of the data booklet

Draw a graph of bond length vs. bond enthalpy (strength) 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. Identify any significant exceptions to this trend

and explain why they occur.

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Key Points

Atoms (generally) form covalent bonds according to the octet rule.

Each covalent bond gives an atom one extra electron

In dative bonds, both the electrons in the bond come from the same atom

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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How many lone pairs and bonding pairs of electrons surround xenon in the XeF4 molecule?

Lone pairs Bonding pairsA. 4 8B. 0 8C. 0 4D. 2 4

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Lesson 3: Structures

Objectives:

Describe and compare the structures and properties of: Allotropes of carbon Silicon and silicon dioxide Ionic compounds

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Marketplace – in four groups Each group needs to produce a learning resource to teach the other

students about their chosen topic.

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.

There will be a test at the end.

Groups should look at the structure, bonding, properties and uses of: 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

Complete the test here

You have 10 minutes

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Alternative to Marketplace Activity1. 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)

2. 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.

3. Produce a Venn Diagram to summarise then similarities and differences between the three main types of bonding

4. 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

Graphene is a recently discovered ‘super-material’

Research its structure and properties and as many potential uses for it as possible

Make sure you have

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Key Points

Carbon (diamond, graphite and fullerenes), silicon and silicon dioxide exhibit giant covalent (macromolecular) structures For example diamond: each carbon is bonded to

exactly four others and so on

Ionic compounds form giant ionic lattices 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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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

Objectives:

Design and conduct an experiment to use the physical properties of compounds to differentiate between them

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Design Challenge You will be provided with samples of 6 unknown substances:

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: 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 Bring laptop to next lesson and make sure you

have installed ACD/Labs ChemSketch (Google it!)

Identify which substance is which out of: Lead Graphite Iodine Sugar Sodium chloride Lead bromide

Fully explain your reasoning

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Key Points

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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State two physical properties associated with metals and explain them at the atomic level.

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Lesson 5: Molecular Shapes

Objectives:

Model, draw, and explain the shapes of molecules with 2-4 charge centres around a central atom

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 Valence Shell Electron Pair Repulsion (aka ‘vesper’)

Pairs of electrons around an atom repel each other – this determines a molecules shape

Pairs of electrons are known as ‘charge centres’ and include both: The electrons in a covalent bond

a double/triple bond only counts as one charge centre! Lone pairs / non-bonding pairs

Example: ethyne The carbon has two charge centres (the C-H bond and the CC 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: Draw as many atoms as

you can in the same plane ‘flat’ on the paper

Use solid wedges to show atoms coming out of the plane of the paper towards you

Use dashed wedges to show atoms going back into the plane of the paper away from you

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Modelling Molecules You need to model molecules with varying numbers of charge centres. For

each one you should: Draw a Lewis structure Draw the structure (in 3D where relevant) Label bond angles and explain them Name the structure

HL and SL: two, three and four charge centres including: CO2, C2H2, BF3, C2H4, SO2, CH4, NH3, H2O, NH4

+, H3O+

HL only: five and six charge centres including: PCl5, SF6, XeF4, PF6

-

How: 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 Negative charge centres repel each other, this determines the

shape of a molecule

Standard shapes: Two charge centres: linear Three charge centres: trigonal planar Four charge centres: tetrahedral Five charge centres: trigonal bipyramidal Six charge centres: octahedral

Lone pairs: 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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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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Lesson 5: Molecular Shapes

Objectives:

Learn to identify and explain the three types of intermolecular forces: 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 The attractive forces between

molecules

It is these that are partially broken during melting, and fully broken during boiling

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 akaTemporary dipole induced dipole forces Random electron movements create a

small, temporary dipole

This induces a similar dipole in a neighbouring molecule

This creates a small attraction between them

These are weak and exist only for the tiniest fraction of a second

Van der Waals forces are present in all molecules

Van der Waals forces: Increase with molecular mass Decrease with the roundness of a molecule

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Dipole-dipole forces akaPermanent dipole forces 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 The end with high charge density is - The end with low charge density is +

Oppositely charged dipoles attract each other.

This is a relatively strong attractive force

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 akaH-bonds 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

Intermolecular forces play an important role in the properties of compounds including:

Melting/boiling point: Stronger intermolecular forces higher m.p./b.p.

Volatility: Stronger intermolecular forces lower volatility

Solubility: like dissolves in like 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

Three types of intermolecular force, from strongest to weakest:

Hydrogen bonds Between N/O/F and H attached to N/O/F

Dipole-dipole 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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Which compound has the highest boiling point?

A. CH3CH2CH3

B. CH3CH2OHC. CH3OCH3

D. CH3CHO

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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 The two best marks from any of your assignments

Data Collection and Processing: 2 x 6 12 marks The two best marks from any of your assignments

Conclusion and Evaluation: 2 x 6 12 marks The two best marks from any of your assignments

Interpersonal Skills: 6 marks Assessed by the Group 4 project

Manipulative Skills: 6 marks Assessed cumulatively over the whole course

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An exemplar

Read through the assessment criteria crib-sheet here

Read through the example internal assessment here

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The topic

The topic for your internal assessment should be:

‘a factor affecting intermolecular forces’

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Defining your research question Things to consider:

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: 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: 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

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 10HL Only

Sigma and Pi Bonds

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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 π-bonding 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

2. CH4

3. O2

4. H2O

1. N2

2. C2H4

3. C3H8

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 p-orbitals 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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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 11HL Only

Sigma and Pi Bonds

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How many sigma (σ) and pi (π) bonds are present in the structure of HCN?

σ πA. 1 3B. 2 3C. 2 2D. 3 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 s- orbitals 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_staff/gutow/Orbitals/N/sp3%20hybrid.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/sp2%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_staff/gutow/Orbitals/N/sp2%20hybrid.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_staff/gutow/Orbitals/N/sp%20hybrid.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

H2O CO2

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).

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

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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 12HL Only

Delocalisation

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Refresh

What is the type of hybridization of the silicon and oxygen atoms in silicon dioxide?

Silicon OxygenA. sp3 sp3

B. sp3 sp2

C. sp2 sp3

D. sp2 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 6-carbon 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 p-

orbitals 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-, CO3

2-, 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 non-

delocalised 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 π-system

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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