Edexcel Unit 1

Cations, X+…

Most singular ions, Na+

Only NH4+ applicable

Anions , X-…

Most molecular/compound ions, NO3

Cl, Br,I, O2-

Soluble Anions

NO3, Cl, Br, I, SO4-2


Insoluble with

Ag+ Pb+2

Insoluble Anions

OH S-2 CO3-2 PO4-3

Soluble with

Na+ KNH4+(except OH) Ba+(only OH)

Relative Atomic Mass
Average mass of one atom of the element as compared to 1/12 mass of one atom of C-12

1 ppm = 1 mg/dm3

= 1 x 10-3 g of substance

1 dm3 of solution

% yield




x 100%

Yield < 100 %

1. XXX reactants not pure

2. Loss of XXX products during transfer

3. XXX side reaction occurs

4. XXX reaction does not reach completion.

Atom Economy

           Molecular Mass of Desired Product         

Sum of Molecular Mass of all reactants/products


x 100

Exothermic Reaction

1. Enthalpy of Products < Enthalp of Reactants

2. -ve Enthalpy Change

3. High Temperature

4. Heat lost to surroundings

5. Bonds formed




Endothermic Reaction

1. Enthalpy of Products > Reactants

2. +ve Enthalpy Change

3. Lower Temperature

4. Gain from surroundings

5. Bonds broken


Thermodynamic Stability
Lower = More Stable
Standard Enthalpy of Formation


1 mole of compound

From its elements

Standard states at 298K and 1 atm


Standard Enthalpy of Combustion

1 mole of substance

Burns completely in oxygen

at 298K and 1 atm

Standard Enthalpy Change of Neutralisation

1 mole of water is formed

From neutralisation of 1 acid by 1 alkali

at 298K and 1 atm

Standard Enthalpy Change of Atomisation

1 mole of separate gaseous atoms

From element in its standard state

at 298K and 1 atm

Bond Enthalpy

Break 1 mole of covalent bond

in gaseous molecules

Experimental method to find Enthalpy Change



– M x C x ?T

– Only use mass of solution and solution’s C as substance/powder etc. have negligible C and M

Possible errors in using calorimeter to measure Enthalpy Change of Neutralisation

1. Heat lost to calorimeter.

2. Assumption density and specific heat capacitry of solution same as water.

Calorimeter – Why Polystyrene cup and not beaker?

1. Reduces heat lost.

2. Inert/ Does not react with most chemicals.

3. Does not absorb much heat.

4. If use beaker, ?H will decrease.

Enthalpy Change of Combustion of fuel

Black solid – Carbon, formed due to incomplete combustion


Experimental of Combustion ; Theoretical


1. Heat lost to surroundings

2. Some heat goes to beaker instead

3. Incomplete combustion due to inadequate O2

4.Condition of products not standard (water vapour, endothermic reaction)



1. Use shields

2. Use copper tin

3. No shields

Why excess substances?

To ensure all XXX have completely reacted




To ensure all XXX have neutralised completely (if neutralisation)

Hess’s Law Cycle

If using enthalpies of formation,

1. Arrows go up

2.  Elements as middle part


If using enthalpies of combustion,

1. Arrows go down

2. Combustion products as middle part

Why XXX reaction cannot be determined directly but calculated through Hess’s Law?


(Alkene to alkane)


CnH2n + H2 -; CnH2n+2

1. Reaction cannot be conducted in lab due to H2 being flammable

2. Ethane etc. may undergo combustion

3. Reaction rate is too slow.

Bonds broken +ve


Bonds formed -ve
3 uses of Table of average standard bond enthalpies

1. Compare strength of bonds

2. Understand struscture and bonding

3. Calculate ?Hf and ?Hreaction
using Hess’s Law

Ensuring accuracy of experiments

1. Constant stirring of water to ensure even temperature

2. Use screen/coppertin/polystyrene board/lid etc to prevent heat loss

3. Use more accurate apparatus (thermometers that read up to 0.1, use pipetter not measuring cylinder…)

4. Take temperature every 30s…

Percentage Error

        Uncertainty of Error     

Value of measurement


x 100


Reliability : repeat experiment and get average value


Accuracy :

1. Use precise instruments

2. Use larger volume/mass…

Which sub-atomic particle would deviate the most?
Electron as it has a lower mass than proton.



1.Same Chemical properties

2.Different Physical properties


1. same amount of electrons/electronic configuration

2. Masses are different

Mass Spectrometer

1. Vaporisation (gaseous atoms to move through instrument)

2. Ionisation (bombarded by high energy electrons and 1 electron is knocked out of atom)

3. Acceleration (Electric field)

4. Deflection (Magnetic field according to m/z ratio)

5. Detection (shows how many ions of each m/z ratio in sample)

Usage of mass spectrometry

1. Archeology – Radiocarbon dating using C-14

2. Geographical – Detect oil composition


s, p, d, f

Atomic Radii


1.Decrease across Period

2.Increase down Group


1. More protons but electrons in same energy level/shells, force of attraction increase


2. More protons but electron in higher energy levels, outermost electrons more shielded, force of attraction decreases

Ionisation energy

Remove 1 mole of electrons

From 1 mole of gaseous atoms

to form 1 mole of gaseous positive ions

Factors of Ionisation energy

1. Distance from nucleus (atomic radii)

2. Nuclear charge (number of protons)

3. Shielding effect (by inner shell electrons)

Trends in first ionisation energy


1. Increase across period

2. Decrease down the group

1. Atomic radius smaller, Nuclear charge increases, Negligible shielding as all in same shell, FOA ^


2. Atomic radius increase but cancelled off by increase in number of protons, but shielding effect increase, FOA decrease

First ionisation energies do not increase smoothly across period
Due to presence of subshells
First Electron Affinity

1 mole of electrons

Added to 1 mole of atoms

In gaseous state

First EA


1. Increase across period


1. Nuclear charge increase, atomic radius smaller,negligible shielding
Why first EA always exothermic and why second EA endothermic?

1. To form an attraction between incoming electrons and nucleus

2. Repulsion between incoming electron and negative ion

Ionic Bonding
Electrostatic force of attraction between oppositely charged ions
Giant Ionic Lattic Structure

1. Neatly arranged

2. Very high melting & boiling pt. – large amount of energy to overcome strong electrostatic attractions and separate ions

3. Hard but brittle – Any dislocation=layers moving=repulsion occurs=splits crystal

4. Good electrical conductivity in molten or aqueous state only – solid=ions held strongly by lattice, liquid/molten=mobile ions & conduction takes place

5. Very high density – higher than water but lower than typical metals

3D arrangement of ions in NaCl

6 Na+ ion around each Cl ion




Cubic structure with alternate Na+ and

Size of cation


1. < size of atom


1. Number of electron shell decreases, higher p/e ratio, Effective nuclear charge increases, remaining e are pulled in closer to nucleus

Size of Anion


1. > Size of Atom

2. Increases down the group

1. Same number of electron shell but p/e ratio decreases, Effective nuclear charge decreases, expansion of electron cloud


2. Number of protons increases, but electrons in higher/outer shell, more shielding, FOA decreases

Isoelectronic ions

Ions with same number of electrons


H Li+ Be2+ B3+   – 2 electrons in each ion

Lattice Energy

1 mole of ionic compound

formed from gaseous ions under standard conditions


1. larger ionic charge = larger lattice energy

2. smaller sum of ionic radii=larger lattice energy

Determination of Lattice energy
Born=Haber Cycle
Chances of formation
higher when amount of energy released high
Polarisation in ionic bonds

Cations – larger charge= higher polarising power

 – smaller redius = higher polarising power


Anions – larger radius= higher polarisability

Why CaI2 more covalent than KI? (Why more difference between theoretical and experimental lattice energy?)

-higher charge, smaller radius

-more polarising

-able to distort electron cloud to a greater extent

Covalent Bonding


Electrostatic force of attraction between the nuclei and shared pair of electrons
Dative Covalent Bonds
Shared pair of electrons which has been provided by one of the bonding atoms
Metallic Structure
A giant lattice of positive metal ions fixed in position and surrounded by a sea of electrons
Metallic Bonding
Force of attraction between the positive metal ion and negative delocalised electrons
Factors on strength of metallic bond

1. Ionic radius

2. Ionic charge

3. Valence electrons (number of delocalised electrons)

Physical properties of Giant Metallic Lattices


1. V.high melting and boiling pt.

2. Good malleability and ductility

3. Good electrical conductivity in both solid and molten states

4. V.high density

1. Strong electrostatic attractions between cations and delocalised electrons

2. Ions and delocalised electrons move around each other, will not break as cations still surrounded by electrons

3. Delocalised electrons free to move and can conduct electricity when a potential difference is applied

4. Higher than water and ionic compound

Why strength of bond between simple molecules/simple atoms weak?
weak Van der Waals forces between molecules/atoms

1. Structural – same molecular formula but different structure


2. Geometrical -restricted rotation of Carbon-Carbond double bond(C=C)

– Both C atoms of C=C must have 2 different groups/atoms

Cis/trans isomers


– must have at least 1 same group of atoms each side

Cis – same group of atoms at same side of C=C


Trans – Groups at different sides


Limits – When all different groups, don’t work

E-Z isomerism

Use mass instead (group with higher mass is used)


E- Different

Z – Same


1. CnH2n+2

2. Only single bonds

3. 4 H bonds to every C atom

4. Saturated Hydrocarbon

Fractional distillation (of crude oil)

1. Higher = not so pure, lower boiling/melting point

2. Middle area purest

3. Lowest= bitumen, Highest=Gas,


Gas;Petrol;Chemicals;Aircraft fuel;Central Heating fuel;Lube;Power Station/Ship fuel;Candles and grease;Road


Use of high temperature or catalyst to break large hydrocarbon to smaller molecules of alkane/alkene


C-C bonds broken


(Can also occur when no oxygen present)


E.g of catalyst zeolite Al2O3

Why short chains high in demand?

1. Small chain of alkane allow for use as fuel

2. Small chain of alkene for use in polymer production

Catalytic Reforming
Process in which straight-chain alkanes form rings or branched chains in presence of high temperature or catalyst(E.g. Pt)
Alkane reactions

1. Combustion of alkanes

2. Free radical substitution (E.g with Cl2) (must have UV)

Bond fission

1. Homolytic

2. Heterolytic


1. CnH2n

2. C=C

3. Unsaturated hydrocarbon

4. More reactive than alkane

5. Can form cis-trans (E/Z) isomers

Nature of C=C
Consists of 1 ? (sigma) and 1 ? (pi) bonds
? bond

1. Overlap of 2 s-orbitals

2. Overlap of 2 p-orbitals (one end to another end)

3. Overlap of one s-orbital and p-orbital

? bond
Side-on overlap of 2 p-orbitals (at 2 points)
? stronger than ?
Sigma bond overlaps are closer to nuclei
C=C – region of high electron density
Electrons more diffused and less firmly held. So easily attacked by electrophiles.
Chemical reactions of alkene

1. Combustion

2. Electrophilic addition

3. Oxidation

4. Polymerisation

Electrophilic addition
Type of reaction in which an electrophile is attracted to the electron density in the C=C and added across the double bond to form one product.

Markovnikov’s rule


(My understanding of it)

The more hydrogen atoms attached to the carbon atoms of the double bond(reactant) in the product, the more likely it is to be the major product. (more stable)
Test for C=C bond (double bond)

Add few drops of Bromine water, Brto alkene.


It will be decolourised from brown to colourless.

Process in which many small molecules (monomers) join together into large molecules (polymers) consisting of repeating units
Addition Polymerisation
alkene monomers are joined together without elimination of any atoms/molecules
General Equation



Common polymers

1. Polyethene – food wrap

2. Polypropene – yogurt tubs

3. Polybutene – Rubber piping

4. Polychloroethen (PVC) – Cable insulation

      -Make flexible using plasticizer to allow chains to slide

5. Polytetrafluoroethene (Teflon) – Non-stick coating on frying pan

6. Polyphenylethen (Polystyrene) – Foam packaging, polystyrene cup

Polymer Problems

1. Use non-renewable resources (from crude oil)

2. High-energy production costs

3. Non-biodegradable – landfill problem

4. Does not burn easily (requires extremely high heat)


1. Reducing

2. Recycling

3. Burn waste (Produces energy)

4. Feedstock (Convert waste to hydrocarbon)

5. Biodegradable polymers (made from natural polymers – starch, cellulose)

Energy Recovery
Reducing energy consumption of polymer manufacture

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