Representing reactions- IGCSE CHEMISTRY WORK SHEETS AND NOTES

Word equationsIf a reaction occurs between magnesium and oxygen, magnesium oxide is produced, here is the word equation for this reaction:

magnesium + oxygen --> magnesium oxide

Some other examples are:

hydrochloric acid + calcium carbonate --> calcium chloride+ carbon dioxide+ water sodium + water --> sodium hydroxide + hydrogen

hydrochloric acid + sodium hydroxide --> sodium chloride + water

Write word equations for the reactions in which the following compounds form from a halogen and another suitable element: hydrogen fluoride, hydrogen chloride, iron III chloride, iron III bromide, sodium chloride, copper chloride.

formulaeThe formula of an element or compound is simply the symbol of each element present and numbers to show how many atoms are present. Carbon dioxide has the formula CO2. This means that it has one carbon atom and two oxygen atoms in each molecule

Use valency to work out the formulae of the following compounds: sodium chloride, potassium bromide, magnesium oxide, calcium sulphide, aluminium nitride, calcium iodide, lithium oxide, aluminium chloride, aluminium sulphide, magnesium nitride.

Calculating relative formula massAdd up the relative atomic mass (found in periodic table) of each atom in the compound.

e.g. Al203 relative atomic masses of Al = 27, O = 16 (found in periodic table).

The formula shows 2 atoms of aluminium and 3 atoms of oxygen so:

formula mass of = (2*27) + (3*16) =54 + 48 = 102

Work out the relative formula masses of the following:

MgO, FeS, O2, H2O, CaBr2, Na2S, CaCO3, NaOH, HCl, (NH4)2SO4.
Relative atomic masses Mg=24, O=16, Fe=56, S=32, Ca=40, Br=80, C=12, Na=23, H=1, Cl=35.5.
Simple balanced equationsIt is possible to write balanced equations for reactions. For example substances such as hydrogen and magnesium combine with oxygen. One method to write them is:Write a word equation first.
Magnesium + oxygen --> magnesium oxide
Write in the formulae of the substances used.
Mg + O2 --> MgO
Balance the equation so that each element has the same number of atoms on each side.
2Mg + O2 --> 2MgO
sodium + oxygen --> sodium oxide
4Na{s} + O2{g} --> 2Na2O{s}
(word equation) hydrogen + oxygen ---> water(formulae)
H2 + O2 -----> H2O(balance)
2H2 + O2 -----> 2H2O


State symbolsThe state symbols are put in a balanced equation to show whether something is a solid, liquid, gas or dissolved in water (aqueous solution). The symbols for these are:
state
Symbol
Solid (s)
Liquid (l)
Gas (g)
Aqueous (aq)
Magnesium + oxygen --> magnesium oxide 2Mg{s} + O2{g} --> 2MgO{s} hydrochloric acid + calcium carbonate --> calcium chloride + carbon dioxide + water2HCl (aq) + CaCO3 ---> CaCl2(aq) + CO2(aq) + H20(l)

Task

Balanced equations and ionic equations

Ionic equations only show ions which change in a reaction and ignore those which do not change.

E.g.word equation
hydrochloric acid + sodium hydroxide --> sodium chloride + water

balanced chemical equation

HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O(l)

ionic equation

H+(aq) + OH-(aq) ---> H2O(l)

E.g. in the electrolysis of sodium chlorideCl-(aq) --> Cl(g) + e-2Cl(g) --> Cl2(g)

Working out formulae from reacting masseselements reacting magnesium chlorine

symbols of elements Mg Clmasses reacting (from experiment)
2.4g 7.1gmolar mass (look up relative atomic 24g/mol 35.5g/molmass in periodic table)amounts (amount = mass/molar mass) 2.4g/24g/mol 7.1g/35.5g/mol = 0.1mol 0.2molratio of atoms (divide by smallest) 1 : 2formula MgCl2



Task :

Work out formulae of compounds formed when the following react:56g of iron and 32g of sulphur (Fe =56, S =32)2g of hydrogen and 16g of oxygen (H=1, O=16)14g of lithium and 16g of oxygen (Li=7)32g of copper and 8g of oxygen (Cu=64)6.4g of copper and 0.8g of oxygen.

Calculating reacting masses using equationsYou can work out ratio of the masses of products and reactants by simply multiplying the number of moles shown in the equation by the formula mass of each substance.

Example 1: What mass of magnesium oxide can be made from 12g of magnesium? Relative atomic masses are Mg =24, O = 16.

equation 2Mg(s) + O2(g) --> 2MgO(s)

formula 2*24 1(16*2) 2(24+16)

masses =48 =32 =80 reacting 48g of Mg forms 80g of MgOmasses 1g of Mg forms 80/48 g of MgO 12g of Mg forms 12*80/48 g of MgO = 20g

Example 2: What mass of magnesium oxide can be made from 12g of magnesium?

equation 2Mg(s) + O2(g) --> 2MgO(s)

amounts 2 moles 1 mole 2 moles masses 2*24 1{16*2} 2{24+16} =48g =32g =80g

so 48g Mg forms 80g MgO 1g Mg forms 80/48 g MgO 12g Mg forms 12*80/48 g MgO = 20g Also note that the ratio of amounts of reactants and products in the equation above can be written as:

Amount of Mg/amount of O2 =2/1 Or Amount of O2/amount of MgO = 1/2


Task : problem C2.08 You decide to travel from London to Delhi for a holiday and hire your own Airbus 319 jet. You fly the 2183 miles to Cairo first and make 9.4 tons of carbon dioxide.


After seeing the Pryramids you fly the 1018 miles to Riyadh making 4.5 tons of carbon dioxide. After a brief stop in the Saudi Capital you fly on the 1900 miles to Delhi making 8.1 tons of carbon dioxide.


Finally you return to London travelling 4171 miles and releasing 22 tons of carbon dioxide. The jet burns the fuel kerosine

C15H32 in the reactionC15H32 + 23O2 --> 15CO2 +16H2O(a)

How much carbon dioxide was released into the atmosphere during the trip?(b) What mass of kerosine was burnt?

(c) What effect does the trip have of the environment?

Alkanes- As level work sheet

Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.

1. One of the terminal steps in the photochemical chlorination of methane is:
A Cl2---------- 2Cl•
B CH3• + CH3• ---CH3 CH3
C CH4 + Cl--- CH3Cl + H+
D CH4 + Cl•-- CH3 + HCl

2. Which one of the following represents a propagation step in the chlorination of methane?
A CH4 + Cl2 -- CH3Cl + HCl
B CH4 + Cl• --- CH3• + HCl
C Cl2 ---- 2Cl•
D CH3• + Cl• --- CH3Cl

3. Which one of the following species is not formed in the photochemical reaction between chlorine and methane?
A H•
B Cl•
C HCl
D CH3CH3

4. Which one of the following is formed in the initiation step in the mechanism of the chlorination of methane?
A chlorine atoms
B hydrogen atoms
C hydrogen chloride
D methyl radicals


5. 10 cm3 of an equimolar mixture of hydrogen and methane were combusted by exploding with 25 cm3 of pure oxygen. All volume measurements were made at 20oC and 1 atmosphere pressure.

The volume of the mixture (in cm3 ) after the explosion was:
A 12.5
B 15
C 17.5
D 20
Section B

Answer all questions

1. Methane reacts with chlorine in the presence of ultraviolet light, the reaction occurring in a series of steps.

(a) The initiation step involves the action of ultraviolet light on chlorine gas.

Write an equation and explain what happens to the chlorine.

equation


explanation

(b) Write two equations for the second step in this mechanism.

1.

2.


(c) Suggest why traces of ethane are found in the product mixture.
2. (a) Kerosene, which contains a mixture of C11 and C12 hydrocarbons, is obtained from crude oil by fractional distillation. Explain the term fractional distillation.

(b) The hydrocarbons from kerosene can be cracked to form other useful products.

(i) Explain the term cracking.

(ii) Using the molecule C11H24, write an equation to illustrate cracking.

(c) Another fraction contains several isomers of C7H16. Draw structures and give systematic names for any two branched chain isomers of C7H16.

Structure (i)
Name
Structure (ii)

(d) Hexane is exploded in the internal combustion engine according to the equation
2C6H14 + 19O2 12CO2 + 14H2O

3.Calculate the molar enthalpy of combustion of C6H14 using the
bond enthalpies given.

kJ mol-1
C-C 346
C-H 413
0-H 453
0=0 490
C=0 740

Alkanes- As level notes edexcel igcse notes

Associated with these alkanes we have alkyl radicals of
general formula CnH2n+1 (symbol R)

e.g. CH3- methyl,
C2H5- ethyl,
C3H7- propyl
This family of alkanes forms an homologous series.

Combustion in a limited and plentiful supply of air.

Combustion

The alkanes are used as fuels and burn in excess air or oxygen producing carbon dioxide and water.
2C2H6 + 7O2 ---- 4CO2 + 6H2O
C3H8 + 5O2 ---- 3CO2 + 4H2O

In general
CnH2n+2 + (3n+1) O2 -nCO2 + (n+1) H2O

In a limited air supply carbon monoxide is produced
2C2H6 + 5O2--- 4CO + 6H2O


Reaction with chlorine or bromine , monohalogenation only (except for methane and chlorine.


Alkanes are generally unreactive. They are saturated and react by substitution.
Halogenation
Chlorine and bromine react with alkanes in the presence of strong sunlight or u.v. light giving a series of products formed by successive replacement of a hydrogen atom by a halogen atom.
R-H + X2 R-X + HX

Mechanism of the photochemical reaction between chlorine and methane viewed as a free radical substitution.

Chlorine reacts explosively with methane in the presence of strong sunlight or u.v. light to give a mixture of products.
CH4 + Cl2 --- CH3Cl + HCl

This is photochemical chlorination.

Mechanism
The mechanism of a reaction is the course believed to be followed by the reactants in combining together and the various stages inv olved in reaching the final products.This is a free radical chain reaction.
Initiation
Cl2 ---2Cl. (a few)

Propogation
Cl. + CH4 CH3. + HCl
CH3. + Cl2 CH3Cl + Cl.
CH3Cl + Cl. CH2Cl. + HCl
CH2Cl. + Cl2 CH2Cl2 + Cl.

Termination
2Cl. Cl2
CH3. + Cl. CH3Cl
2CH3. C2H6


Isomerism in Organic Compounds

Structural isomerism for aliphatic compounds containing up to six carbon atoms, to include branched structures. (Cyclic structures excluded)

Isomerism

Strutural isomerism occurs when 2 or more compounds have the same molecular formula but different structural formula. (ie same number of atoms but bonded together differently).

Draw and name all the structural isomers of C4H10, C5H12 and C6H14.

Environmental problems associated with spillage and combustion of alkane fuels.

Unreactive nature of alkanes towards electrophiles and nucleophiles.

Hydrocarbons- work sheet/ igcse chemistry /As chemistry

1. The systematic name of the following compound

CH3-CH2-C( CH3)Br-CH3

is
A 2-bromo-2-methylbutane
B 3-bromo-3-methylbutane
C 2-bromopentane
D 3-bromopentane

2. 25 cm3 of hydrocarbon was exploded with 100 cm3 of oxygen (an excess). After cooling the volume of gas remaining was 62.5 cm3. On adding concentrated potassium hydroxide and shaking the volume reduced to 12.5 cm3. Which one of the following is the molecular formula of the hydrocarbon?
A CH4
B C2H4
C C2H6
D C3H6

HYDROCARBONS

Petroleum as a source of hydrocarbons by fractional distillation and cracking.

A knowledge of how these processes are carried out, including the names and nature of the fractions obtained, is expected.(Reforming to benzene derivatives is not required in this module.)

Environmental problems associated with spillage and combustion of hydrocarbons, including global warming and ozone depletion.

HYDROCARBONS
(alkanes, alkenes, benzene.)

Sources

1. Petroleum.

Petroleum deposits were formed by the action of pressure and temperature on marine life sediments, mainly from the fatty acid constituents, under the catalytic action of various rocks and acid clays.

Crude petroleum is a complex mixture of gaseous, liquid and solid hydrocarbons such as alkanes , cycloalkanes, aromatics (benzene) and some alkenes. Also present are some compounds of oxygen, nitrogen and sulphur.

It has no uses in its raw form so to provide useful products its components must be partly separated and if necessary modified. The fundamental process of refining is primary distillation.

Refining

The function of an oil refinery is to manufacture from crude oil those quantities of the oil products required by consumers. This is carried out by the use of various physical and chemical processes.

Refinery gas

( 1 - 2 per cent of crude oil) contains hydrocarbons that are gases at normal temperatures. It includes the alkanes with one to four carbon atoms in their molecules, with methane as a major component. The main use of refinery gas is as a gaseous fuel.

Gasoline

( 15 - 30 per cent) has is a complex liquid mixture of hydrocarbons containing mainly C5 - C10 compounds whose boiling points range from 40oCto 180oC. The major use of gasoline is as a fuel in internal combustion engines. A considerable part proportion of this fraction is used to produce chemicals by cracking.

Kerosene

( 10 - 15 per cent) consists mainly of C4 to C 12 hydrocarbons, with boiling point from 162 to 250 degrees C. It is used as a fuel in jet engines and for domestic heating. It can be cracked to produce extra gasoline.

The diesel oil or gas oil

(15 - 20 per cent) containing C 13 - C25 compounds, boils between to 120 - 350 degrees C. It is used in diesel engines where the fuel is ignited by compression instead of by a spark. And also for industrial heating purposes. It can also be cracked to produce extra gasoline.

Residue

(40 - 50 per cent) boils above 350 degrees C and is a highly complex mixture of non-volatile hydrocarbons. Most of it is used as fuel oil in large furnaces such as those in power stations or big ships. Proportion of it is used to make lubricating oils and waxes. Both these contain C 26 - C 40 carbons

Distillation

Crude oil is fractionally distilled to give 4 main fractions.
C1-C4
refinery gases
methane, ethane, propane, butane
C8-C16
light distillates
petrol, aviation fuel, kerosene, benzene
C17-C20
middle distillates
heating oil, diesel, feedstock for cracking
C21+
residue
paraffin wax, lubricating oil, petroleum jelly, bitumen

2. Coal Tar

When coal is carbonised (burnt in the absence of air) one of the products is a viscous black liquid called coal tar. This is distilled into five main fractions. The light oil fraction (boil pt. up to 170oC) is a source of benzene.

Cracking as a source of alkenes and shorter chain alkanes.

Cracking

In this process larger molecules are broken down into smaller ones, either by high temperature and pressure (thermal cracking) or by a catalyst (catalytic cracking).

Thermal Cracking

When alkanes are heated to high temperatures their molecules vibrate strongly enough to break and form smaller molecules. One of these molecules is usually an alkane. Reducing chain length generally results in unsaturation. Such reactions are known as cracking

e.g. C8H18 C5H12 + CH3CH=CH2

octane pentane propene
Thermal cracking is generally used for cracking residues to middle distillates.

Catalytic Cracking

By using a catalyst, cracking can be made to occur at fairly low temperature. This is known as catalytic cracking.

Catalytic cracking is the most important source of petrol and raw materials for the chemical industry. Heavier fractions can be cracked to produce extra gasoline. Cracking tends to produce branched-chain rather than straight-chain alkanes, so the gasoline produced this way has a high octane rating. Processes similar to cracking can be used to convert low-grade gasoline to high grade fuel.

The catalysts are usually natural clays and synthetic alumina/ silica mixtures (Al2O3/SiO2).

Isomerisation

This involves breaking up straight chainalkanes and reassembling them as branched chain isomers.
Both of these processes are important in the production of unleaded gasoline.



Catalytic Reforming

Reforming involves converting straight chain alkanes into ring molecules such as arenes and cycloalkanes.

Benzene C6H6 and other aromatic compounds can be made by passing petrol vapour over a heated platinum catalyst.
500oC/15 atm
C6H14 -,--- C6H6 + 4H2
hexane Pt catalyst benzene

The u.s.a. obtains about half its benzene in this way.

ENERGETICS - SUMMARYNOTES

1 ENERGETICS

Simple treatment (including calculations) of the Born-Haber cycle for the halides of Group I and
(Lattice enthalpy will be regarded as the enthalpy of lattice breaking)

When you have finished this section you should be able to:

Explain and use the term ‘lattice enthalpy’ as a measure of ionic bond strength.
Construct Born-Haber cycles to calculate the lattice enthalpy of a simple ionic solid

e.g NaCl, MgCl2, using relevant energy terms (enthalpy changes of formation, ionisation energy, enthalpy of atomisation and electron affinity.

Explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of a lattice energy.

Lattice Enthalpy :

Bond enthalpies provide a measure of the strength of covalent bonds. Ionic bonding is an electrostatic attraction between oppositely charged ions.

The attraction acts in all directions, resulting in a giant ionic lattice containing many ions. For ionic compounds the corresponding enthalpy is the lattice enthalpy (also called the lattice energy).

Lattice enthalpy indicates the strength of the ionic bonds in an ionic lattice.

The standard molar lattice enthalpy is the energy required to convert one mole of a solid ionic compound into its constituent gaseous ions under standard conditions.

e.g. NaCl (s) Na+ (g) + Cl- (g)

Born-Haber cycle
This is an application of Hess’s Law and can be used to calculate lattice energies.
Lattice enthalpies cannot be determined directly by experiment and must be calculated indirectly using Hess’s Law and other enthalpy changes that can be found experimentally.

The energy cycle used to calculate a lattice enthalpy is the Born-Haber cycle.
The basis of a Born-Haber is the formation of an ionic lattice from its elements.
In general for an ionic compound a Born-Haber cycle can be written as:

M+ (g) + X- (g)

M+ (g) + X (g) ΔHe.a.

ΔHdiss.
M+ (g) + ½ X2 (g)

ΔHI.E. ΔHlatt.
M (g) + ½ X2 (g)

ΔHsub.
M (s) + ½ X2 (g)

ΔHFθ

M+X- (s)

According to Hess’s Law
ΔHlatt. = (-ΔHFθ) + ΔHsub. + ΔHI.E. + ΔHdiss. + ΔHe.a.

ΔHfθ = enthalpy of formation of MX (s)
ΔHsub. = enthalpy of sublimation of M (s)
ΔHI.E. = ionisation energy
ΔHdiss. = dissociation energy of X2 (g)
ΔHe.a.= electron affinity of X (g)
ΔHlatt. = lattice energy

N.B. The actual figures may be positive or negative and are simply substituted in the above equation.

Consider the reaction between sodium and chlorine to form sodium chloride.
Na (s) + ½ Cl2 (g) NaCl (s)

The reaction can be considered to occur by means of the following steps:

Vapourisation of sodium
Na (s) Na (g) ΔHsub.
The standard enthalpy of sublimation or vaporisation is the enthalpy change when one mole of sodium atoms are vaporised. This is an endothermic process and can be determined experimentally.

Ionisation of sodium
Na (g) Na+ (g) + e- ΔHI.E
The standard enthalpy of ionisation is the energy required to remove one mole of electrons from one mole of gaseous atoms. This is endothermic and can be determined by spectroscopy.

· Dissociation of chlorine molecules
Cl2 (g) 2Cl (g) ΔHdiss.
The standard bond dissociation enthalpy is the energy required to dissociate one mole of chlorine molecules into atoms (i.e. to break one mole of bonds). This is also endothermic and can be determined by spectroscopy.

Ionisation of chlorine atoms

Cl (g) + e- Cl- (g) ΔHe.a.
The electron affinity of chlorine is the energy released when one mole of gaseous chlorine atoms accepts one mole of electrons forming one mole of chloride ions.
Reaction between the ions

Na+ (g) + Cl- (g) NaCl (s) -ΔHlatt.
This is the reverse of the lattice energy. The standard lattice enthalpy is the energy absorbed when one mole of solid sodium chloride is separated into its gaseous ions. It has a positive value and cannot be determined experimentally.


A Born-Haber cycle can be drawn:
Na+ (g) + Cl- (g)

ΔHe.a.
Na+ (g) + Cl (g)

ΔHI.E.
Na (g) + Cl (g) ΔHlatt.

ΔHdiss.
Na (g) + ½ Cl2 (g)

ΔHsub. ΔHFθ
Na (s) + ½ Cl2 (g) NaCl (s)

Applying Hess’s Law
ΔHsub. + ΔHI.E. + ΔHdiss. + ΔHe.a. - ΔHlatt. – ΔHfθ = 0

Calculate the lattice enthalpy of sodium chloride given
ΔHfθ (NaCl) = -411 kJ mol-1
ΔHsub. (Na) = 108.3 kJ mol-1
ΔHI.E. (Na) = 500 kJ mol-1
ΔHdiss. (Cl) = 121 kJ mol-1
ΔHe.a. (Cl) = -364 kJ mol-1
Answer ΔHlatt. = +776 kJ mol-1


Exercise 1
Draw Born-Haber cycles for each of the following ionic compounds and calculate their lattice enthalpies.
(Note : ΔHat. of an element is the energy required to form one mole of gaseous atoms from the element.)


Solubility of ionic compounds is usually governed by its lattice energy. In general the higher the lattice energy the lower the solubility.
(See enthalpy of solution below)
A comparison of calculated and theoretical lattice energies gives an indication of the degree of covalent character in an ionic compound. The greater the difference between the two values the more covalent the compound.

The close agreement between the theoretical and experimental values for the alkali metal halides provides strong evidence that the simple ionic model of a lattice, composed of discrete spherical ions with an even charge distribution, is a very satisfactory one.

For the silver halides the theoretical values are about 15% less than the experimental values based on the Born-Haber cycle. This indicates that the simple ionic model is not very satisfactory.

When there is a large difference in electronegativity between the ions in a crystal , as in the case of the alkali metal halides then the ionic model is satisfactory. However as the difference in electronegativity gets smaller, as in the case of the silver halides, the bonding is stronger than the ionic model predicts.

The bonding in this case is not purely ionic but intermediate in character between ionic and covalent. The ionic bonds have been polarised (Fajans rules) giving some covalent character.


Exercise 2

The figures below give a list of lattice energies in kJ mol-1. Try to find as many patterns and trends in the figures as you can.

RbF -779 CaI2 -2038
BeF2 -3456 CaCl2- 2197
BaI2- 1841 MgCl2- 2489
MgBr2- 2416 KCl -710
CaBr2 -2125 NaF- 915
CsI- 607 LiF- 1029
KBr -671 MgI2 -2314
BaF2 -2289 LiBr- 804
CsBr -644 RbI- 624
LiI -753 SrBr2- 2046
BeI2 -2803 NaBr- 742
LiCl- 849 SrCl2- 2109
NaI -699 BeBr2- 2895
BeCl2- 2983 KF -813
CsCl -676 BaBr2- 1937
KI- 643 CaF2- 2583
MgF2 -2883 NaCl -776
RbCl- 685 SrF2 -2427
SrI2- 1954 RbBr -656
CsF -735 BaCl2- 2049



Trends in lattice enthalpy explained in terms of ionic radius and charge.

Consider the ionisation of an ionic solid MX.
MX (s) Mn+ (g) + Xn- (g)
The ease of separation of the ions and hence the lattice energy is determined by the size of the ions and their charge.


Effect of ionic size
As the ionic radius of both Mn+ and Xn- the lattice energy decreases. The attractive force between the ions decreases and they become easier to separate.
e.g. LiBr 804 kJ mol-1 BeCl2 2983 kJ mol-1
NaBr 742 MgCl2 2489
KBr 671 CaCl2 2197
RbBr 656 SrCl2 2109
CsBr 644 BaCl2 2049

As we descend both Groups I and II the lattice energies become less positive.
For any given metal the lattice energy also decreases in passing from the fluoride to the iodide.
e.g. NaF 915 kJ mol-1 SrF2 2427 kJ mol-1
NaCl 776 SrCl2 2109
NaBr 742 SrBr2 2046
NaI 699 SrI2 1954
This is due to an increase in ionic size from F- to I- which increases the internuclear distance. There is a corresponding decrease in attractive force and hence lattice energy.
When the internuclear distances are about equal, as for RbF and LiI for example, then the lattice energies are almost equal.


Effect of ionic charge
As the charge on Mn+ increases there is a greater attractive force between the ions and lattice energies increase. In addition, the decrease in size of Mn+ with increasing charge increases the attractive force between the ions and also increases the lattice energy.
The ionic radius of the Na+ and Ca2+ ions are very similar. However the lattice energy of CaCl2 is about 3 times that of NaCl.

NaCl 776 kJ mol-1 CaCl2 2197 kJ mol-1

This is due to the increased charge on the metal ion giving greater electrostatic attraction.
In general Group II halides have a lattice energy about three times that of the equivalent Group I halide.
Beryllium halides have considerable covalent character and the lattice energies are bigger than expected.

Exercise 3
What would be the effect on lattice energy of increasing the charge on Xn- ? (i.e. forming a Group VI compound rather than a Group VII compound).
Describe and explain the trends.

For comparable interionic distances the lattice energy would be bigger for X2- ions compared with X- ions. This is because X2- ions exert a stronger electrostatic field compared to X- ions due to the increased charge.
For comparable interionic distances the lattice energy is approximately four times higher when M2+ X2- ions are involved comared to M+ X-.

Exercise 4
Describe and explain the trends in hydration energy.

As level Isomerism notes- work sheet- As chemistry work sheet

1. Which compound may exist as cis and trans isomers?

A CH3CH=CHCH3

B CH2=CHCl

C CH= CH2

D CH3 CH2


2. What is the total number of isomeric alkenes of formula C4H8?
A 2
B 3
C 4
D 5

3. Which one of the following molecules can exist as cis-trans isomers?
A (CH3)2C=CHCH3
B CH3CH2CH=CHCH3
C H2C=CHCH2CH2CH3
D CH3CH(CH3)CH=CH2

Isomerism in organic compounds

Isomerism in Organic Compounds

Structural isomerism for aliphatic compounds containing up to six carbon atoms, to include branched structures and position of the C=C double bonds in alkenes. (Cyclic compounds excluded).

Isomerism (greek isos, equal; meros, parts) occurs when there are two or more compounds (called isomers) with the same molecular formula but different arrangements of their atoms.

Isomers have different physical and chemical properties but the differences may be great or small depending on the type of isomerism.
There are two main classes of isomerism;


(i) structural isomerism

(ii) stereoisomerism

which are themselves sub-divided.

ISOMERISM

structural isomerism stereoisomerism
chain positional functional group geometric optical
isomerism isomerism isomerism isomerism isomerism

Structural isomerism

Structural isomers are molecules with the same molecular formula but with different structural arrangements of the atoms.

Chain isomers

This occurs where isomers have different arrangements of the carbon chain. There is only one alkane corresponding to each of the formulae CH4, C2H6 and C3H8.



For butane C4H10 two arrangements are possible.

n- butane
butane methylpropane
b.pt. 273K b.pt. 261K

As the number of carbon atoms increases the number of possible isomers increases rapidly.

Formula
Number of structural isomers
C5H12-3
C6H14-5
C7H16-9

Positional isomers

This occurs when isomers have the same carbon skeleton but the functional group is in different positions in the molecule.
propan-1-ol
propan-2-ol

Exercise 1
(a) Draw and name the five structural isomers of C6H14.

(b) Write the structural formulae of all the alcohols of molecular formula C4H10O. Name the isomers.


Stereoisomerism : cis-trans isomers for compounds containing one C=C bond, the energy barrier to rotation in these compounds.

Stereoisomerism
Stereoisomers have identical molecular formulae, and the atoms are linked together in the same order, but have different relative positions in space.
The two types of stereoisomerism are
(i) Geometric (or cis-trans ) isomerism
(ii) Optical isomerism [See Module 4.5]

· Geometric isomers
Geometric or cis-trans isomers exist because the π bond of the C=C bond prevents free rotation .

H H Cl H
C=C C=C
Cl Cl H Cl
cis-1,2-dichloroethane trans-1,2-dichloroethane

Cis-trans isomerism can also occur in inorganic complexes about a single bond with square planar or octahedral structures.

e.g. diaminedichloro platinum (II)


Exercise 2
1. Draw and name all the structural isomers, including geometric isomers, of C4H8.

AS LEVEL -The Periodic Table

The organisation of elements in the Periodic Table according to their proton numbers and electronic structures. The terms group and period. The trends in the physical properties across the period sodium to argon limited to melting points, electrical conductivity, first ionisation energies and atomic radii.
Group VII (fluorine, chlorine, bromine and iodine)
Practical work restricted to chlorine, bromine and iodine and their compounds.
Trends within the group limited to colour, physical state, melting and boiling points, atomic and ionic radii, first ionisation energies, bond energies of halogen molecules, hydrogen halides and carbon-halogen bonds; electronegativies.

The halogens are a group of reactive non-metals, which are essentially similar to each other with only gradual changes as the atomic number increases.

Physical properties of the halogens

Property
Fluorine -F
Chlorine -Cl
Bromine -Br
Iodine -I
Colour
Pale yellow
Pale green
Red/brown
Black
Physical state
Gas
Gas
Liquid
Solid


They are all p-block elements with a simple molecular structure consisting of covalently bonded diatomic molecules, X2.

There are only weak Van der Waals forces between the molecules. The strength of the forces increases as the number of electrons (and Mr) in the molecule increases.

F2<> Br2 > I2 as the atoms get larger and the attraction of the nucleus for the shared electrons decreases (electronegativity decreases).
There is a slight tendency to metallic character with increasing atomic number. The halogens complete their octet by gaining one electron forming a halide ion, X- (see electron affinity values) or by sharing one electron.

Solubility in water and non-aqueous solvents eg hexane.

Solubility of the elements
All three elements are only slightly soluble in water because of the relatively strong hydrogen-bonding between the water molecules, which does not exist between the halogen molecules
i.e. solvent-solvent attractions > solute-solvent attractions > solute-solute
attractions.

Cl2> Br2>I2
solubility decreasing


They are soluble in non-polar organic solvents such as toluene and TCE.
(Why?)





Chemical trends: reactivity with hydrogen, sodium and phosphorus.

All the halogens are oxidising agents and combine readily with metals and non-metals.

Reaction of halogens with elements
Hydrogen
The halogens combine enthusiastically with hydrogen, the vigour of the reaction decreasing from fluorine to iodine.
H2 (g) + X2 (g) 2HX (g)
Fluorine reacts explosively even in the dark at –200 oC.
Chlorine reacts explosively in sunlight, or slowly in the dark below 200 oC.
Bromine reacts above 200 oC and at lower temperatures with a platinum catalyst.
Iodine reacts to form an equilibrium mixture
H2 (g) + I2 (g) ⇌ 2HI (g)

Metals
The halogens combine readily with most metals forming the metal halides.
The vigour of the reaction decreases from fluorine to iodine.
Group I and II halides are ionic.

2Na (s) + Cl2 (g) 2Na+Cl- (s)
Mg (s) + Cl2 (g) Mg2+2Cl- (s)

The halides of Group III are predominantly covalent.

2Al (s) + 3Cl2 (g) 2AlCl3 (s)

Non-metals
The elements react directly with many non-metals the oxidising power decreasing from fluorine to iodine.
The elements combine directly with phosphorus, the oxidation state of the product depending on the oxidising power of the halogen:

2P (s) + 5Cl2 (g) 2PCl5 (s)
2P (s) + 3Br2 (l) 2PBr3 (l)

Reactions of the elements illustrated by use of chlorine gas (or chlorine water), bromine water and aqueous iodine (in potassium iodide) with water, aqueous alkalis, other halides in solution and iron (II) and iron (III) ions as appropriate. Disproportionation.

Reaction of halogens with water
Fluorine and chlorine can oxidise water. Fluorine oxidises water to oxygen.
2F2 (g) + 2H2O (l) 4HF (aq) + O2 (g)

Chlorine reacts slowly with water forming hydrochloric acid and chloric(I) acid. This reaction involves disproportionation:- a change in which one particular molecule, atom or ion is simultaneously both oxidised and reduced.

reduction

Cl2 (g) + H2O (l) HCl (aq) + HClO (aq) (chlorine water)
o.n. 0 -1 +1
oxidation

Chlorine water contains chloric (I) acid HClO (aq), (hypochlorous acid). This is a weak acid which ionises to give the chlorate (I) ion ClO-, (hypochlorite ion). The hypochlorite ion is a powerful disinfectant and bleach.

Bromine disproportionates in a similar way but to a lesser extent.
Iodine has a very low solubility in water.



Reaction of halogens with aqueous sodium hydroxide.
Chlorine reacts faster with dilute sodium hydroxide than with water.
When chlorine is added to cold dilute alkali it disproportionates to chloride and chlorate(I).
(i)
reduction

Cl2 (g) + 2NaOH (aq) NaCl (aq) + NaOCl (aq) + H2O
o.n. 0 -1 +1
oxidation


( 2OH- + Cl2 Cl- + OCl- + H2O )


(ii) In hot concentrated alkali, if the solution is warmed to 70oC, the chlorate(I) disproportionates further to chlorate(V).

reduction

3NaOCl (aq) 2NaCl (aq) + NaClO3 (aq)
o.n. +1 -1 +5
oxidation

If chlorine is bubbled directly into hot conc. alkali then

(iii) reduction

3Cl2 (g) + 6NaOH(aq) 5NaCl (aq) + NaClO3 (aq)
o.n. 0 -1 +5
oxidation

( 6OH- + 3Cl2 5Cl- + ClO3- + 3H2O )

For bromine, both reactions (i) and (ii) are fast at 15oC.
For iodine, decomposition of IO- occurs rapidly at 0oC so it is difficult to prepare NaIO free from NaIO3.
NaClO is a mild antiseptic (Milton).
NaClO3 is a powerful weed killer.

Displacement reactions of the halogens
Since they are very electronegative, all the halogens are oxidising agents. As the group is descended their oxidising power decreases.

Therefore chlorine oxidises bromide ions to bromine and iodide ions to iodine.
These are displacement reactions.

Cl2 (g) + 2Br- (aq) Br2 (l)+ 2Cl- (aq)
(colourless) (yellow/orange)

Cl2 (g) + 2I- (aq) I2 (s) + 2Cl- (aq)
(colourless) (red/brown)

Bromine oxidises iodide to iodine

Br2 (g) + 2I- (aq) I2 (s) + 2Br- (aq)

Iodine does not oxidise any of the others.

Other oxidising reactions of the halogens
The trend in oxidising power is illustrated by the compounds formed by iron when it combines directly with the halogens.
Fluorine and chlorine form iron(III) fluoride and iron(III) chloride respectively.
2Fe (s) + 3F2 (g) 2FeF3 (s)

Bromine forms both iron(II) bromide and iron(III) bromide.
Iodine is too weak an oxidising agent and only forms iron (II) iodide.
Fe (s) + I2 (g) 2FeI2 (s)

Aqueous solutions of chlorine, bromine and iodine oxidise iron (II) to iron (III).
Cl2 (aq) + 2Fe2+ (aq) 2Cl- (aq) + 2Fe3+ (aq)

Iodine is so weak an oxidising agent that iron (III) ions oxidise iodide ions to iodine.

2Fe3+ (aq) + 2I- (aq) 2Fe2+ (aq) + I2 (s)


















Thermal stability of hydrogen halides related to bond enthalpies. The relative strength of the acids, HF, HCI, HBr and HI.

Thermal stability of hydrogen halides
The thermal stability of the hydrogen halides decreases as the group is descended. This is in keeping with the trend in bond enthalpies

The size of the halogen atom increases from fluorine to iodine; therefore the bond length increases and the bond enthalpy decreases.
Hydrogen fluoride and hydrogen chloride are stable to heating. Hydrogen bromide decomposes on strong heating.
2HBr (g) H2 (g) + Br2 (g)
Hydrogen iodide decomposes on gentle heating. If a hot wire is dipped into hydrogen iodide gas violet clouds of iodine are produced.
2HI (g) H2 (g) + I2 (g)

Hydrogen halides as acids
The dry hydrogen halides are not acidic and do not affect litmus paper. The hydrogen halides dissolve readily in water forming acid solutions.
HX (g) + H2O (l) H3O+ (aq) + X- (aq)

The acid strength increases in the order
HF << HCl < HBr < HI
An aqueous solution of HF is weakly acidic. This is because the H-F bond is very strong and there is strong hydrogen bonding between the HF molecules.
The rest are all strong acids.

Ionic halides. The identification of halide ions in solution by use of silver ions followed by aqueous ammonia. The effect of light on silver halides. Presence of halide ions in sea water. The reaction of solid halides with concentrated sulphuric acid to illustrate the relative reducing ability of halides ions and hydrogen halide. The effects of fluoridation of public water supplies on dental health and an appreciation of the debate between public health policy and practice and the rights of the individual.


Reaction of the halide ions in solution, X-(aq)
Most metal halides are soluble except lead and silver halide. Therefore solutions of lead and silver ions are used to test for the presence of halide ions in solution.

Reagent
F- (aq)
Cl- (aq)
Br- (aq)
I- (aq)
Pb(NO3)2 (aq)

Pb2+(aq) + 2X-(aq) PbX2(s)
White precipitate of PbF2
White precipitate of PbCl2
Cream precipitate of PbBr2
Yellow precipitate of PbI2
AgNO3 (aq)

Ag+ (aq) + X- (aq) AgX (s)
No reaction AgF soluble in water
White precipitate AgCl
Cream precipitate AgBr
Yellow precipitate AgI
Solubility of silver halide in
(a) dil. NH3 (aq)
(b) conc. NH3
(c) dil.HNO3 (aq)


Exercise 2
Write an equation for the reaction of sodium chloride solution with
(a) lead nitrate solution and





(b) silver nitrate solution followed by the addition of ammonia.





Halide ions in sea water

Sea water contains over 3% of dissolved chlorides (mainly sodium, potassium, calcium and magnesium) and large solid deposits are found where inland seas have undergone evaporation. Bromine is also found in sea water as the bromides of sodium, potassium and magnesium. Iodine is present in sea water at a concentration of less than one part per million. Some seaweeds and sponges extract this iodine during growth and their ash contains about 0.5% by mass of iodine as iodides.


Reaction of the solid halides with conc. sulphuric acid, H2SO4

When concentrated sulphuric acid is added to a sodium halide the first product is fumes of the hydrogen halide HX, because each of these compounds is more volatile than sulphuric acid.
NaCl (s) + H2SO4 (1) HCl (g) + NaHSO4 (s)

NaBr (s) + H2SO4 (1) HBr(g) + NaHSO4 (s)

However conc. sulphuric acid is also a strong oxidising agent and will oxidise
HBr Br2 and HI I2, but not HF and HCl.



oxidised

2HBr(g) + H2SO4 Br2 + SO2 (g) + 2H2O

reduced


Similarly 2HI(g) + H2SO4 I2 + SO2(g) + 2H2O

Therefore conc. sulphuric acid cannot be used for the preparation of
HBr(g) and HI(g).
However conc. phosphoric(V) acid, H3PO4, can be used for the preparation since it is relatively non volatile and a poor oxidising agent.

NaBr (s) + H3PO4 (1) HBr (g) + NaH2PO4
NaI (s) + H3PO4 (1) HI (g) + NaH2PO4

Reagent
Fluoride
Chloride
Bromide
Iodide
Conc. H2SO4

X- + H2SO4 HX (g) + HSO4-
HF (g) colourless, pungent, corrosive gas
HCl (g)
formed
HBr (g) + a little Br2
A little HI (g) but mainly I2
Conc. H3PO4

HF (g)
HCl (g)
HBr (g)
HI (g)

Exercise 3
Make notes on the uses of the halogens and their compounds, with particular reference to the fluoridation of public water supplies and the effects on dental health.

KEY SKILLS

AS LEVEL CHEMISTRY - PERIODIC TABLE NOTES

The Periodic Table

The organisation of elements in the Periodic Table according to their proton numbers and electronic structures. The terms group and period. The trends in the physical properties across the period sodium to argon limited to melting points, electrical conductivity, first ionisation energies and atomic radii.

Group VII (fluorine, chlorine, bromine and iodine)


Practical work restricted to chlorine, bromine and iodine and their compounds.
Trends within the group limited to colour, physical state, melting and boiling points, atomic and
ionic radii, first ionisation energies, bond energies of halogen molecules, hydrogen halides and carbon-halogen bonds; electronegativies.

GROUP VII Halogen

(Halogens : restricted to chlorine, bromine and iodine)
Valid deductions may be expected about other elements in the group.

Physical properties of halogens, limited to colour and physical state at room temperature.

The halogens are a group of reactive non-metals, which are essentially similar to each other with only gradual changes as the atomic number increases.

Physical properties of the halogens

Property
Fluorine
F
Chlorine
Cl
Bromine
Br
Iodine
I
Number of protons
(atomic number)
9
17
35
53
Outer electron configuration


2s22p5
3s23p5
3d104s24p5
4d105s25p5


Atomic radius/ nm
0.064
0.099
0.111
0.128
Ionic radius/ nm


0.133
0.181
0.196
0.219

Melting point / oC
-220
-101
-7
114

Boiling point / oC

-188
-34
58
183
Bond energy /

kJ mol-1
158
242
193
151
Electron affinity/ kJ mol-1

-361
-388
-365
-332
Standard electrode potential/ V
+2.87
+1.36
+1.09
+0.54
Electronegativity
4.00
2.85
2.75
2.20
Oxidation states
- 1
-1,1,3,5,7
-1,1,3,5,7
-1,1,3,5,7
Standard enthalpy of formation of NaX/ kJ mol-1
-573
-414
-361
-288
Standard lattice enthalpy of NaX/ kJ mol-1
902
771
733
684


They are all p-block elements with a simple molecular structure consisting of covalently bonded diatomic molecules, X2.

o o o o
o o o
o X o X o X X
o o o o
There are only weak Van der Waals forces between the molecules. The strength of the forces increases as the number of electrons (Mr) in the molecule increases.

F2<> Br2 > I2 as the atoms get larger and the attraction of the nucleus for the shared electrons decreases (electronegativity decreases).

There is a slight tendency to metallic character with increasing atomic number. The halogens complete their octet by gaining one electron forming a halide ion, X- (see electron affinity values) or by sharing one electron. Fluorine is restricted to an oxidation state of -1 but the remaining elements have empty d orbitals and can promote electrons to give oxidation states of +1, +3, +5 and +7.

They are all oxidising agents and combine readily with metals and hydrogen.

Chlorine is a greenish-yellow gas.
Bromine is a red-brown volatile liquid.
Iodine is a black shiny solid, which sublimes on heating to produce a purple vapour.

Solubility in water and non-aqueous solvents eg hexane.

Chemical trends: reactivity with hydrogen, sodium and phosphorus.

Reactions of the elements illustrated by use of chlorine gas (or chlorine water), bromine water and aqueous iodine (in potassium iodide) with water, aqueous alkalis, other halides in solution and iron (11) and iron (111) ions as appropriate.

Disproportionation.
Reaction of halides with elements

Metals

The halogens combine readily with most metals forming the metal halides.
The vigour of the reaction decreases from chlorine to iodine.
Group I and II halides are ionic.


2Na (s) + Cl2 (g) 2Na+Cl- (s)
Mg (s) + Cl2 (g) Mg2+2Cl- (s)

The halides of Group III are predominantly covalent.

2Al (s) + 3Cl2 (g) 2AlCl3 (s)

Non-metals

The elements react directly with many non-metals the oxidising power decreasing from chlorine to iodine.

The elements combine directly with phosphorus, the oxidation state of the product depending on the oxidising power of the halogen.

2P (s) + 5Cl2 (g) 2PCl5 (s)
2P (s) + 3Br2 (l) 2PBr3 (l)

Solubility of the elements

All three elements are only slightly soluble in water because of the relatively strong hydrogen-bonding between the water molecules, which does not exist between the halogen molecules
i.e. solvent-solvent attractions > solute-solvent attractions > solute-solute
attractions.
Cl2> Br2>I2
solubility decreasing

They are soluble in non-polar organic solvents such as toluene and TCE.
(Why?)


Chlorine reacts slowly with water forming hydrochloric acid and chloric(I) acid. This reaction involves disproportionation:- a change in which one particular molecule, atom or ion is simultaneously both oxidised and reduced.

reduction

Cl2 (g) + H2O (l) HCl (aq) + HClO (aq) (chlorine water)
o.n. 0 -1 +1
oxidation

Bromine and iodine disproportionate in a similar way but to a lesser extent.
Reaction of chlorine with aqueous sodium hydroxide.
Chlorine reacts faster with dilute sodium hydroxide than with water.
When chlorine is added to cold dilute alkali it disproportionates to chloride and chlorate(l).
(i) reduction

Cl2 (g) + 2NaOH (aq) NaCl (aq) + NaOCl (aq) + H2O
o.n. 0 -1 +1
oxidation

( 2OH- + Cl2 Cl- + OCl- + H2O )

(ii) In hot concentrated alkali, if the solution is warmed to 70oC, the chlorate(I) disproportionates further to chlorate(V).

reduction

3NaOCl (aq) 2NaCl (aq) + NaClO3 (aq)
o.n. +1 -1 +5
oxidation

If chlorine is bubbled directly into hot conc. alkali then

(iii) reduction

3Cl2 (g) + 6NaOH(aq) 5NaCl (aq) + NaClO3 (aq)
o.n. 0 -1 +5
oxidation

( 6OH- + 3Cl2 5Cl- + ClO3- + 3H2O )

For bromine, both reactions (i) and (ii) are fast at 15oC.
For iodine, decomposition of IO- occurs rapidly at 0oC so it is difficult to prepare NaIO free from NaIO3.
NaClO is a mild antiseptic (Milton).
NaClO3 powerful weed killer.

Thermal stability of hydrogen halides related to bond enthalpies. The relative strength of the acids, HF, HCI, HBr and HI.

Ionic halides. The identification of halide ions in solution by use of silver ions followed by aqueous ammonia. The effect of light on silver halides. Presence of halide ions in sea water. The reaction of solid halides with concentrated sulphuric acid to illustrate the relative reducing ability of halides ions and hydrogen halide. The effects of fluoridation of public water supplies on dental health and an appreciation of the debate between public health policy and practice and the rights of the individual.
Reaction of the halide ions in solution, X-(aq)
Most metal halides are soluble except lead and silver halide. Therefore solutions of lead and silver ions are used to test for the presence of halide ions in solution.

Reagent
F- (aq)
Cl- (aq)
Br- (aq)
I- (aq)
Pb(NO3)2 (aq)

Pb2+(aq) + 2X-(aq) PbX2(s)
White precipitate of PbF2
White precipitate of PbCl2
Cream precipitate of PbBr2
Yellow precipitate of PbI2
AgNO3 (aq)

Ag+ (aq) + X- (aq) AgX (s)
No reaction AgF soluble in water
White precipitate AgCl
Cream precipitate AgBr
Yellow precipitate AgI
Solubility of silver halide in
(a) dil. NH3 (aq)
(b) conc. NH3
(c) dil.HNO3 (aq)




soluble
soluble
insoluble

insoluble soluble
insoluble

insoluble insoluble
insoluble

Effect of sunlight

White ppt. turns purple/grey
Cream ppt. turns green/ yellow

No effect

Exercise 2
Write an equation for the reaction of sodium chloride solution with
(a) lead nitrate solution and



(b) silver nitrate solution followed by the addition of ammonia.

A LEVEL CHEMISTRY - PERIODIC TABLE

1.8 The Periodic Table
Section A
For each of the following questions only one of the lettered responses (A-D) is correct.
Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the response sheet.

1.B Which one of the following equations does not represent a reaction of chlorine under suitable conditions?
A Cl2 + H2O HCl + HOCl
B 2Cl2 + CH4 CCl4 + 2H2
C 3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O
D Cl2 + 2NaOH NaCl + NaOCl + H2O


2.A When the hydrogen halides dissolve in water acidic solutions are formed. Which has the highest pH (assuming all are equal concentrations)?
A Hydrogen fluoride
B Hydrogen chloride
C Hydrogen iodide
D Hydrogen bromide


3. In which one of the following changes has chlorine been oxidised?
A 3Cl2 + 2Fe 2FeCl3
B Cl2 + I2 2IC1
C Cl2 + 2KBr Br2 + 2KCl
D Cl2 + 2KOH H2O + KClO + KCl


4. In which one of the following sequences are the oxides of the elements classified as basic, amphoteric and acidic respectively?
A Na, Mg, Al
B Na, K, S
C K, Al, P
D S, P, Mg


5. Which one of the following is true of the halogens as the Group is descended from chlorine to iodine?
A The atomic radius decreases.
B The colour of the element lightens.
C The melting point of the element increases.
D The oxidising power of the element increases.
5. Which one of the following equations represents a reaction of chlorine?
A Cl2 + NaOH NaOCl + HCl
B Cl2 + Fe FeCl2
C Cl2 + H2O HCl + HOCl
D Cl2 + CH4 CH2Cl2 + H2

SECTION B

1. The group in the Periodic Table headed by fluorine is called the halogen group.

(a) Using your knowledge of the group predict the colour and physical state of astatine.
Black / solid

(b) When chlorine gas is bubbled through a solution of potassium iodide a redox reaction occurs.

(i) What is observed in this reaction?
The colour of the gas disappears / Solution turns colourless to dark brown


(ii) What is meant by a redox reaction?
A reaction in which one or more electrons / is~are transferred from one substance to another


(iii) Explain fully the changes undergone (if any) by the potassium and iodide ions.
Potassium ion : no change (1)/ iodide ion : loses electron to form iodine (1) [2]
(c)
(i) The presence of iodide ions in an aqueous solution of sodium iodide can be detected using silver nitrate solution and ammonia solution. Describe fully how you would carry out this procedure experimentally, stating any changes you observe. Write equations for any reactions which occur.
Add dilute nitric acid followed by silver nitrate solution to the iodide solution/ A pale yellow / precipitate / of silver iodide / forms. This is insoluble in ammonia solution. / AgNO3 + NaI = AgI + NaNO3 (1)

(ii) State what differences you would observe if the sodium iodide solution was replaced by sodium chloride solution. Write equations for any reactions which occur.
With sodium chloride a white precipitate of silver chloride forms. This is soluble in ammonia solution

2. Sodium bromide reacts with concentrated sulphuric acid at room temperature to give four products. Place ticks (√) in four of the boxes to show which substances are formed.
√ if the substance is formed
bromine

hydrogen

hydrogen bromide

hydrogen sulphide

sodium hydrogensulphate

sulphur

sulphur dioxide

3. The diagram below shows chlorine gas being passed through a dilute
solution of potassium iodide. The upper layer is a hydrocarbon solvent.
(a) (i) What is the most important safety precaution to take when carrying
out the experiment?
The colour of the gas disappears / Solution turns colourless to dark brown

(ii) Write the ionic equation for the reaction between chlorine and potassium iodide and explain the redox reactions taking place in terms of electron transfer.

With sodium chloride a white precipitate of silver chloride forms. This is soluble in ammonia solution

(b) When the aqueous layer is shaken with the hydrocarbon most of the iodine dissolves in the upper layer.

(i) What is the colour of iodine in potassium iodide solution?

(ii) What does the greater solubility in the hydrocarbon suggest about the bonding in iodine?
The colour of the gas disappears / Solution turns colourless to dark brown

(iii) The percentage composition of the elements in the solvent is:

element
% composition by mass
carbon
92.3
hydrogen
7.7

Calculate the empirical formula of the solvent.

(c) If the hydrocarbon layer is shaken with aqueous sodium thiosulphate the colour disappears.

(i) Write an equation for the reaction between iodine and sodium thiosulphate.

(ii) If insufficient thiosulphate solution were added traces of iodine are left which are not visible to the eye. Name a reagent that could be added to detect the iodine and state the colour produced.

ATOMIC STRUCTURE

1.1 ATOMIC STRUCTURE
Ionisation Energy
Section A

For each of the questions only one of the lettered responses (A-D) is correct.

Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.

1. Which one of the following has the largest first ionisation energy?
A carbon
B fluorine
C neon
D nitrogen


2. The second ionisation energy for calcium is represented by the equation
A Ca(s) Ca2+ (g) + 2e-
B Ca(g) Ca 2+ (g) + 2e-
C Ca+(s) Ca2+ (g) + e-
D Ca+(g) Ca2+ (g) + e-


3.






















4. The first six ionisation energies of an element Y are as follows:
786, 1580, 3230, 4360, 16 000, 20 000 kJ mol-1
Which one of the following statements is consistent with these data.
A Y is in Group I of the Periodic Table.
B The outer electronic configuration of an atom of Y is ns2np2
C Y forms a chloride with the molecular formula YCl
D The chemistry of Y and its compounds is similar to that of aluminium and its compounds.


5. In which one of the following series of elements is there the smallest variation in the value of the first ionisation energy?
A Li to F
B F to I
C K to Br
D Sc to Zn


6. Which one of the following processes requires the largest amount of energy?
A He (g) He+ (g) + e-
B Ne (g) Ne+ (g) + e-
C Na (g) Na+ (g) + e-
D Ca (g) Ca+ (g) + e-


7. Which one of the following statements is correct?
A The first ionisation energy of oxygen is greater than that of nitrogen.
B The second ionisation energy is always greater than the first ionisation energy for a chosen element.
C The first ionisation energy of an element. M. is the energy associated with the process M(s) M+ (aq) + e-
D Ionisation energy increases with increasing atomic number.


8. Which one of the following sets of ionisation energies corresponds to an element in Group II?

1st 2nd 3rd 4th 5th
A 580 1800 2700 11600 14800
B 1520 2700 3900 5800 7200
C 790 1600 3200 4400 16100
D 740 1500 7700 10500 13600


9. What frequency of radiation in Hz is required to ionise helium?

He He+ + e- ΔH = +2370 kJ mol-1

A 1.69 x 10-16
B 9.48 x 10-10
C 1.05 x 109
D 5.93 x 10


10.The first ionisation energy for carbon is greater than it is for sodium. One of the factors responsible is that the
A nuclear charge on carbon is greater
B shielding provided by the inner quantum shells of sodium is greater
C number of electrons in the outer quantum shell of carbon is greater than in sodium
D outer quantum shell is further from the nucleus in carbon atoms than in sodium atoms


11. The ground state electronic configurations of five elements are shown below. For which element would you expect the value of the first ionisation energy to be the greatest?
1s 2s 2p
A
B













C
D












12.An element has successive ionisation energies of 99, 1800, 14800 and 21000 kJ mol-1. To which group of the Periodic Table does the element belong?
A I
B II
C III
D IV


13. The first four ionisation energies of an element Z are 738, 1451, 7733 and 10541 kJ mol-1. Which one of the following ions is most likely to be formed when Z reacts with fluorine?
A Z+
B Z2+
C Z3+
D Z4+

14. The graph represents the variation in the first ionisation energy with atomic number.

The elements indicated by letters P, Q, R and S are all
A alkali metals
B halogens
C noble gases
D transition metals



15. The graph below shows the first six successive ionisation energies, Ie, of an element. In which Group of the Periodic Table is the element found?

A Group 11
B Group IV
C Group V
D Group VI


SECTION B

Answer all questions in the spaces provided

1. Here is some information concerning element X, of atomic number 31. In its natural state, it consists of a mixture of two isotopes XA and XB.

Isotope
Isotopic mass
Percentage abundance
XA
69.0
60.2
XB
71.0
39.8

Its first four ionisation energies are 580, 2000, 3000 and 6200 kJ mol-1.

(a) Calculate the value for the relative atomic mass of X correct to three significant figures.
Ar (X) = (69.0 x 60.2 + 71.0 x 39.8)/100 = 69.8
[3]

(b) Write down the electronic configuration of element X in its ground state.
X = 1s22s22p63s23p63d104s24p1 [1]

(c) In which group of the Periodic Table is X placed?
Group III [1]

(d) Explain why:
(i) The difference between the first and second ionisation energies is greater than that between the second and third ionisation energies.
p electron in higher energy level and better shielded. Therefore easier to remove. [2]

(ii) The difference between the third and fourth ionisation energies is much larger than that between the other successive energies.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]










2. The first five ionisation energies, in kJ mol-1, of two elements Q and R are shown below:

First
Second
Third
Fourth
Fifth
Q
600
1800
3000
12000
15700
R
520
5500
7100
9500
13500

(a)
(i) In which groups of the Periodic Table would you expect to find elements Q and R? explain your reasoning for element Q.
[3]

(ii) Explain why the element directly below R in the Periodic Table would be expected to have a smaller ionisation energy than R.
[3]

(b) Using element R define the term second ionisation energy and write an appropriate equation.
[3]



3. (a) The first ionisation energies for a sequence of elements are shown in the graph:
(i) State whether the first ionisation energy of potassium would be more or less than that of sodium. Explain your choice.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]

(ii) Explain the shape of the graph.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]

(b) The successive ionisation energies of two elements A and B are listed below:
units are kJ mol-1.

1st 2nd 3rd 4th 5th 6th
A 420 3100 4301 5891 7999 9512
B 589 1098 4811 6498 8080 10482


(i) Which group in the periodic table does element A belong to?
Explain briefly your reasoning.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]

(ii) What is the likely charge on an ion of element B? Explain your reasoning.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
4th electron removed from a shell nearer to the nucleus and at a lower energy level than the third. Therefore much more difficult to remove.
[2]


4. The graph below shows the first ionisation energies of the elements
from hydrogen to sodium.


(a) What is meant by the term first ionisation energy?



[2]
(b) (i) Explain why there is an overall rise in first ionisation energy
in going from lithium to neon.





[2]

(ii) Explain why the first ionisation energy falls markedly from helium to lithium.





[2]

(iii) Explain why there is a fall in the first ionisation energy from nitrogen to oxygen.





[2]

(iv) Explain why the first ionisation energy of sodium is less than that of lithium.





[2]

Electronic Structure

1.1 ATOMIC STRUCTURE
Electronic Structure
Section A
For each of the questions only one of the lettered responses (A-D) is correct.

Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.

1. Which one of the following represents the electronic arrangement of the outer shell of the chlorine atom (atomic number, 17)?


3s


3p

A

­¯

­¯
­¯
­
B

­

­¯
­¯
­¯
C

­¯

­
­
­
D

­

­¯
­
­

2. Which one of the following is the electronic configuration of the bromine atom?
A 1s22s22p63s23p63d104s14p6
B 1s22s22p63s23p63d104s24p7
C 1s22s22p63s23p63d104s24p5
D 1s22s22p63s23p63d104s24p6

3. To which one of the following energy levels (denoted by n) does the electron in a hydrogen atom return when the emission spectrum is in the ultraviolet region;
A n= 1
B n = 2
C n = 3
D n = 4

4. Which one of the following does not represent the electronic configuration of an atom in its ground state?
A 1s2 2s2 2p3
B 1s2 2s2 2p4
C 1s2 2s2 2p6 3s1
D 1s2 2s2 2p6 3d1



5. Which one of the following is the maximum number of atomic orbitals having a principal quantum number of two?
A 2
B 4
C 8
D 9

6. Which one of the following contains no unpaired electrons in the ground state?
A Be
B N
C Si
D F

7. Which one of the following is the electronic structure of a metal with a maximum oxidation state of +3?
A 1s2 2s2 2p6 3s1
B 1s2 2s2 2p6 3s2 3p1
C 1s2 2s2 2p6 3s2 3p6 3d10 4s2
D 1s2 2s2 2p6 3s2 3p4

8. Which one of the following statements regarding electronic orbitals is correct?
A Each p-orbital can hold a maximum of six electrons.
B The 3p-orbitals have a higher energy level than the 3s-orbital.
C The three 3p-orbitals have slightly different energy levels.
D The 1s-orbital has the same size and shape as the 2s-orbital.

9. The number of electrons in the 3d orbital of the atom of atomic number 23 is
A 2
B 3
C 4
D 5






10. Which one of the following correctly represents the electronic structure of the named atom in its ground state?
A B C D
2p

2s

1s




2p

2s

1s



2p



2p

2s

1s










2s











1s


Lithium
Boron Carbon Nitrogen

11. The vanadium atom (atomic number 23) in its ground state has the electronic configuration:
A 1s2 2s2 2p6 3s2 3p6 3d3 4s2
B 1s2 2s2 2p6 3s2 3p6 3d2 4s3
C 1s2 2s2 2p6 3s2 3p6 3d1 4s2 4p2
D 1s2 2s2 2p6 3s2 3p6 3d2 4s2 4p1

12. Which one of the following does not have the electronic structure
1s2 2s2 2p6 3s2 3p6 3d10 4s24p6?
A Kr
B S2-
C Se2-
D Sr2+

13. The graph below shows the first ionisation energies for a series of elements whose atomic numbers increase in sequence. Which one could represent nitrogen?



D
1st ionisation C
energy
B
A

atomic number


14. Which one of the following is the electronic configuration of a titanium atom?
A 1s22s22p63s23p64s24p2
@B 1s22s22p63s23p63d24s2
C 1s22s22p63s23p63d'4s'
D 1s22s22p63s23p'4s'4p'

15. Which one of the following compounds contains two ions with different electronic configurations?
A CaCl2
B MgF2
C MgO
D NaCl

16. Which graph represents a plot of energy, E, against principal quantum number (energy level), n, for an electron in a hydrogen atom?






17. Which one of the following atoms contains no unpaired electrons in its ground state?
A Beryllium
B Boron
C Carbon
D Lithium

18. Which one of the following sub-shells does not exist?
A 2p
B 2d
C 3p
D 4d

19. Which one of the following represents the electronic structure of the chloride ion in its ground state?

20. Which one of the following represents the shape of a p-orbital?B
















Section B

1. An outline of the periodic table is shown below.
(a) Indicate on it the positions of the s block, p block and d block elements. [2]





s P d





(b) Write the electronic configuration for
(i) a potassium atom
1s22s22p63s23p64s1
(ii) a sulphur atom
1s22s22p63s23p4 [2]

(c) Write the empirical formula for potassium sulphide indicating the charges on the ions.
K+2S2- [1]


2.
(a) Write the electronic configuration of an iron atom in the ground state
K+2S2- [1]

(b) Explain the term ground state.
K+2S2-
K+2S2-
K+2S2- [1]

3. The diagram below shows the first four energy levels in a sodium atom. Label these levels and, using arrows to represent electrons, show the electronic structure of sodium in the ground state.








[3]


4. Beryllium occurs to a small extent in the earth's crust. It is a steel-grey metal, which is extremely light. Beryllium is used for "windows" in X-ray apparatus because it has the lowest stopping power per unit mass thickness of all suitable construction materials.
Calculate the energy of X-rays of frequency 3 x 1017 Hz passing through a "beryllium window".
K+2S2-
K+2S2-
K+2S2- [2]

5. The Periodic Table is divided into s, p and d blocks. Using the symbols s, p and d complete the table below.
element
block
beryllium

manganese

phosphorus

[2]

ATOMIC STRUCTURE

1.1 ATOMIC STRUCTURE
Spectroscopy
Section A

For each of the questions only one of the lettered responses (A-D) is correct.

Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.

1. The flame colour associated with potassium salts is
A Green
B Lilac
C Yellow
D Red

2. Which one of the following statements about the emission spectrum of hydrogen is correct?
A The number and/or position of lines in the spectrum cannot be influenced by external factors.
B The complete spectrum occurs in the visible region.
C The value for the ionisation energy of hydrogen can be determined from the spectrum.
D All the lines in the spectrum correspond to electron transitions from higher energy levels to the ground state.


3. Which one of the following diagrams represents the visible region of the atomic hydrogen spectrum?

A
Energy

B

Energy

C
Energy

D

Energy

4. The atomic spectrum of hydrogen shows
A a continuum in which individual lines are not apparent
B a series of lines which get closer together as the frequency increases
C equally spaced lines not confined to the visible spectrum
D equally spaced lines in groups separated by a continuum in the visible spectrum

5. In the line spectrum of atomic hydrogen in which region of the electromagnetic spectrum will lines not appear?
A infra-red
B ultraviolet
C visible
D X-ray

6. In which one of the following does the flame coloration correspond to the ion?
A Mg2+ yellow
B Sr2+ green
C Ba2+ white
D Ca2+ red


Section B

1. The presence of sodium ions in a sample of sea salt can be determined by flame colours.

(a) What is the flame colour associated with sodium compounds?
[1]

(b) The most prominent line in the visible spectrum of the atomic sodium spectrum has a frequency of 5.085 x 1014 Hz. Calculate the energy, E, of radiation with this frequency.







[2]

2. The Sun largely consists of a mixture of hydrogen and helium, the presence of each being detected by spectroscopy. The line emission spectrum of atomic hydrogen in the ultraviolet region of the electromagnetic spectrum is shown below.



frequency
(a) Explain why this spectrum consists of lines which are converging.







[5]

(b) Explain how the ionisation energy of atomic hydrogen can be calculated from the spectrum.




[2]


3. State the flame coloration for the following chlorides.

Metal chloride
colour
barium chloride

lithium chloride

potassium chloride

[3]

ATOMIC STRUCTURE- MASS SPCTROMETER

1.1 ATOMIC STRUCTURE
Mass Spectrometry
Section A

For each of the questions only one of the lettered responses (A-D) is correct.

Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.

1. Using a mass spectrometer, it is possible to determine the number of
A protons in an atom
B energy levels in an atom
C isotopes of an element
D neutrons in an atom

2. The accurate relative isotopic masses of five isotopes are
11H = 1.0078 21H = 2.0141 126C = 12.000 147N = 14.0031 168O = 15.9949
Using a high resolution mass spectrometer, a certain gas was found to have a relative molecular mass of 28.0172. The gas could be
A 147N2
B 126C2 11H4
C 21H 126C 147N
D 126C2 21H2

3. The mass spectrometer trace for naturally occurring magnesium is shown below. Assuming that all three peaks relate to ions with one positive charge, what is the relative atomic mass for magnesium?
A 24.2
B 24.3
C 24.4
D 24.7






4. The isotopic composition of a certain element X is 80% 24X, 10% 25X and 10% 26X. The relative-atomic mass of X is
A 24.25
B 24.30
C 24.33
D 24.67

5. The diagram below shows the mass spectrometer trace of the substance X in the region of relative mass 10 to 20 units.









Which of the following substances is X most likely to be?
A O2
B NH3
C H2O
D CH4

6. Which one of the following could not be obtained using a mass spectrometer?
A The number of electrons in an isotope of manganese
B The number of isotopes present in a sample of chlorine
C The RMM of a sample of cocaine
D The mass of an isotope of uranium

9 The mass spectrum of an element is as shown below.

Assuming each fragment has unit charge, what is the relative atomic mass of the element?
A 206.25
B 207.00
C 207.77
D 208.00


SECTION B
1. The mass spectrum of naturally occurring rubidium, Rb, is shown below.
From this spectrum calculate the relative atomic mass of rubidium.

100



55



0 m/z 85 87




[3]
Ar = 100 x 85 + 55 x 87 = 8500 + 4785 = 13285 = 85.71
155 155 155

2.
(a) An organic compound has the composition by mass.

Element
Percentage
Hydrogen
4.1
Carbon
24.2
Chlorine
71.7

(i) Calculate the empirical formula of the compound.







(ii) The mass spectrum of the compound showed the molecular ion peak at a mass/charge ratio of 98. Suggest a structural formula for the compound.






(iii) Suggest why smaller peaks may be found at 100 and 102.
[2]
3. A mixture of 21H2 and 8135Br2 was analysed in a mass spectrometer. The following pattern of lines due to singly-charged ions was obtained.
2 4 81 83 162
mass number on instrument scale

Suggest which ions give rise to each of these lines.

Mass number
Ion responsible
2
21H+
4
21H2+
81
8135Br+
83
21H8135Br+
162
8135Br2+
[3]


4. Beryllium occurs to a small extent in the earth's crust. It is a steel-grey metal, which is extremely light. Beryllium is found in nature as 9Be.

(i) How could you show that naturally occurring beryllium consisted only of 9Be?
[2]

(ii) Draw the structure of 9Be in terms of the constituent particles of the atom.







[2]

4. (a) Complete the table below to show the relative masses and charges of a proton, a neutron and an electron. [3]


Relative mass
Relative charge
Proton


Electron


Neutron



(b) Describe the process by which particles are ionised in a mass spectrometer. [2]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………

(c) Give two reasons why particles must be ionised before being analysed in a mass spectrometer. [2]
Reason 1 ……………………………………………………………………………………………………………………
Reason 2 ………………………………………………………………………………………………………………….

(d) A sample of boron contains 20% by mass of 10B and 80% by mass of 11B. Calculate the relative atomic mass of boron in this sample. [2]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………

(e) Compound X contains only boron and hydrogen. The percentage by mass of boron in X is 81.2%. in the mass spectrum of X the peak at the largest value of m/z occurs at 54.
(i) Use the percentage by mass data to calculate the empirical formula of X. [3]
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………………………
(ii) Deduce the molecular formula of X. [1]
………………………………………………………………………………………………………………………………………

AS level chemistry NOTES ATOMIC STRUCTURE

1. Atomic Structure

Electrons, protons and neutrons as the constituent particles of the atom. Their location in the atom, their relative masses and charges. Atomic number, mass number and isotopes.

When you have finished this section you should be able to:
· Describe the properties of protons, neutrons and electrons in terms of their relative charge and relative mass ;
· Understand the importance of these particles in the structure of the atom ;
· Define the terms atomic number, Z and mass number, A ;
· Use values for atomic number and mass number to calculate the number of protons and neutrons in the nucleus ;
· Explain the existence of isotopes
· Use isotopic symbols to describe the composition of the nucleus.

All atoms are electrically neutral. The number of electrons in the shells is the same as the number of protons in the nucleus. The mass is made up almost entirely from the masses of the protons and neutrons. The masses of the proton and neutron are virtually identical.

Properties of sub-atomic particles

Proton (p)
1
+1
Neutron (n)
1
0
Electron (e)
0 (1/1837)
-1

Evidence for particles
Electrons
J.J. Thompson (1897) experiments with cathode ray tubes. He discovers a beam of rays emitted from the cathode when an electric current passes through a gas at low pressure. The rays are deflected by electric and magnetic fields and consist of a stream of electrons.

Protons
Electric discharges through gases at low pressure produces a stream of particles from the anode. Work with electric and magnetic fields shows them to be positively charged. Hydrogen gives the lightest particles, which are assumed to be protons.
Neutrons
Chadwick (1932) bombards beryllium with a-particles producing fast moving particles, which are not affected by electric or magnetic fields. The particles are neutrons.
Relative sizes of atoms and nuclei
Atoms consist of a very small, dense nucleus, around which the electrons circulate in a comparatively large volume. The volume of the nucleus is about 10-44 m3 and it is composed of two different types of particles, protons and neutrons. These are known collectively as nucleons.
An atom is small, but its nucleus is smaller still. The radius of an atom is of the order of 10-10 m, but the radius of a nucleus is of the order of 10-15 m.

Do the next exercises will help you to appreciate the difference in size.

Exercise 1
Suppose a football, diameter 22 cm, is scaled up so that it becomes as big as the earth, diameter 13000 km.
Calculate whether an atom of diameter 0.32 nm (3.2 x 10-10 m) will become as big as:
A a pin head, diameter 1mm
√B a 1p coin, diameter 1.9 cm

C a football, diameter 22cm
D a weather balloon, diameter 1.8 m

Exercise 2
If the nucleus of an atom were scaled up to the size of a pin head (say 1 mm diameter), how big would the atom be?
100 m (about the length of a football field


Since the mass of an atom is concentrated in its nucleus, the nucleus must be extremely dense. Estimate how dense it is by doing the next exercise.


Exercise 3
For atoms of elements at the beginning of the Periodic Table the volume of the nucleus, VN, is given by:
VN = 1.73 x (relative atomic mass) x 10-45 m3
Use this expression to calculate the density of the sodium nucleus:
(a) in kg m-3 [9.60 x 1017 kg m-3]
(b) in tonnes cm-3 [9.60 x 108 tonnes cm-3]
(Remember that 1 mol of sodium atoms weighs 23.0 g and contains 6.02 x 1023 atoms)

Exercise 4
Calculate:
(a) the volume occupied by a sodium atom (radius 1.86 x 10-10 m)
(b) the fraction of the volume occupied by the nucleus.
(Hint: assume that both the atom and its nucleus are spheres with volume given by 4πr3/3 )

Most of your body is empty space too. If all the spaces between the nuclei were squeezed out, you would be only half as big as a flea, although you would weigh the same.
With all this empty space why does any object appear solid? The electrons in an atom move very rapidly around the nucleus. The electrons effectively form a shield around the nucleus, marking the limits of the atom’s volume and making it seem solid.



Atomic number, Mass number and Isotopes
Atomic number and mass number give us important information about an atom and are particularly useful in distinguishing one isotope of an element from another.

Atomic number
The atomic number (Z) of an atom is the number of protons in the nucleus.


Mass number
The mass number (A) is the total number of particles in the nucleus.


Isotopes
Isotopes are atoms of an element that have different numbers of neutrons in the nucleus i.e. they have the same atomic number but different mass numbers.
e.g. 3517Cl and 3717 Cl.


Exercise 5
The table shows the mass number and number of neutrons in the nucleus, for four atoms, W, X, Y and Z.

W X Y Z
Mass number 36 39 40 40
Neutrons in nucleus 18 20 21 22

a) Write down the atomic numbers of the four atoms.W=18; X=19; Y=19;Z=18
b) Which of the four atoms are isotopes of the same element? W and Z;X and Y
c) Use your Periodic Table to write isotopic symbols (e.g. 2713Al) for the four atoms.
3618W; 3919X; 4019Y; 4018Z

Relative atomic mass, relative isotopic mass and relative molecular mass. The carbon-12 standard. The use of the mass spectrometer to obtain accurate atomic masses. (Details of the workings of the mass spectrometer are not required). Deduction of Relative Molecular Mass from a molecular ion peak. (Limited to ions with single charges).

When you have finished this section you should be able to:
· Calculate the masses of coins relative to a chosen standard ;
· Express masses in a variety of units ;
· Define the terms relative atomic mass (Ar), relative isotopic mass and relative molecular mass (Mr) in terms of carbon-12 ;


Relative atomic mass
Atoms are so small that their masses, expressed in grams, are difficult to work with. Some examples are listed in Table 1 below

Table 1
Element
Average mass of atom g
H
1.67355 x 10-24
He
6.64605 x 10-24
Li
1.15217 x 10-23
C
1.99436 x 10-23
O
2.65659 x 10-23
Na
3.81730 x 10-23
Ar
6.63310 x 10-23
U
3.95233 x 10-22

The mass of an atom expressed as relative atomic mass (Ar) is much more manageable.

Exercise 6
Collect as many British coins of each kind (1p, 2p, 5p, 10p) as you can. Weigh a group of each kind to the nearest 0.01 g and calculate the average mass of each of the denominations to the nearest 0.001 g.
Enter your results in Table 2.

Table 2
Coin
Number of coins
Total mass g
Average mass g
1p
12
43.12
3.593
2p
9
63.49
7.054
5p
8
45.16
5.645
10p
10
110.68
11.068

You can calculate the relative mass of each of the coins using

Relative mass of coin = average mass of coin
mass of standard


Fill in column 1 of Table 3

Exercise 7
Table 3

Coin
1 Average mass g
Relative mass
2 Mass OCU
3 Mass CCU
4 Mass SCU
1p
3.593
1.000


2p
7.054



5p
5.645



10p
11.068




(a) Define the unit of mass, the OCU (for one-penny coin unit).
Let one OCU equal the average mass of a one penny coin.
1.00 OCU = g

(b) Calculate the relative mass of each type of coin on the OCU scale using
Relative mass of coin = average mass of coin
mass of OCU
Fill in column 2

Exercise 8
(a) Define a second unit of mass, the SCU (for silver coin unit) .
Using the data in Table 1 calculate

Average mass of 5p coin = g
1.000 SCU = average mass of 5p coin
1.000 SCU = g

(c) Calculate the relative mass of each type of coin using the expression:
Relative mass of coin = average mass of coin
Mass of SCU

Record the values in column 4 of Table 1.

Exercise 9
According to the Royal Mint, the mass of a newly minted 2p coin is 7.128 g. We define a second unit, the CCU (copper coin unit), as one half the mass of a newly minted 2p coin.
1.000 CCU = ½ x 7.128 g
1.000 CCU = 3.564 g
Using the defined value for the CCU, calculate the relative masses for column 3.

The relative atomic mass scale
You have now completed a series of exercises using coins to illustrate how relative mass changes as the choice of standard changes. Now you will do a similar exercise using masses of atoms instead of coins, where you calculate the relative atomic masses on the hydrogen, oxygen and carbon-12 scales

Exercise 10
Use the values in Table 1 to calculate atomic masses relative to
(a) hydrogen,
(b) oxygen,
(c) carbon-12
in a similar way to that in which you calculated relative masses of coins.

Complete Table 4.
Some values are included as a check.
(The mass of an atom of carbon-12 = 1.99252 x 10-23 g).

Table 4

Element
Relative atomic mass (Ar)
H scale
O scale
12C scale
H
1.00000
1.00794
1.00790
He
3.97123
4.00276
4.00260
Li
6.88459
6.93924
6.93897
C
11.9169
12.0115
12.01110
O
15.8740
16.0000
15.9994
Na
22.8096
22.9907
22.9898
Ar
39.6349
39.9496
39.9480
U
236.164
238.039
238.030

In most of you’re A level work, you use relative atomic masses expressed to three significant figures (e.g. He = 4.00, O = 16.0, U = 238). To this degree of precision, the oxygen scale and the carbon-12 scale can be regarded as the same, but you should not use the hydrogen scale as it differs so much from the others.



RELATIVE ATOMIC MASS (Ar)
Atoms are so light that their actual masses are not used. Instead, each atom is compared to a standard atom. The atom chosen as standard is the most common isotope of carbon, carbon-12.
The carbon-12 atom has a mass of exactly 12.0000 units and all other atoms are given a mass relative to the carbon-12 standard.
For example, a magnesium atom is twice as heavy as a carbon-12 atom.

The relative atomic mass of an element

relative atomic mass of an element = mass of one atom of the element
(1/12) x mass of one atom of carbon-12



RELATIVE MOLECULAR MASS (Mr)
The Relative Molecular Mass of a compound is the sum of the relative atomic masses of all the atoms in a molecule of a compound.
For example: find the Relative Molecular Mass of sulphuric acid.

Formula H 2SO4
2 atoms of H = 2.00
1 atom of S = 32.0
4 atoms of O = 64.0
TOTAL = 98.0

R.M.M. sulphuric acid is 98.

RELATIVE FORMULA MASS
The Relative Formula Mass of an ionic compound equals the sum of the Relative Atomic Masses of all the atoms in a formula unit of the compound.

For example: find the Relative Formula Mass of magnesium chloride.

Formula MgCl2

1 atom of Mg = 24.0
2 atoms of Cl = 71.0
TOTAL = 95.0

R.F.M. magnesium chloride is 95.0.


Find the Relative Formula Mass of hydrated copper sulphate crystals, CuSO4.5H2O.
1 atom of copper Cu = 64.0
1 atom of sulphur S = 32.0
10 atoms of hydrogen H = 10.0
9 atoms of oxygen O = 144
TOTAL = 250
R.F.M. hydrated copper sulphate is 250.The Mass Spectrometer

When you have finished this section you should be able to:
· Understand the principles of a simple mass spectrometer, limited to ionisation, acceleration, deflection and detection ;
· Identify peaks on a simple mass spectrum and use them to calculate the relative abundance and masses of ions ;
· Calculate the relative atomic mass of an element from (a) a mass spectrum, (b) percentage abundance data ;
· Sketch a mass spectrum, given relevant data.

The mass spectrometer is used to obtain accurate atomic masses by measuring the mass and relative abundance of the isotopes of an atom.

1. The sample is vaporised – atoms must be in a gaseous state.
2. Positive ions are formed. Atoms are bombarded by electrons and positive ions are formed .
X(g) X+(g) + e-
3. The positive ions are accelerated by an electric field. The slits restrict the ions to a narrow beam.
4. A magnetic field deflects the ions depending on their mass/charge ratio. Ions with high m/z ratio are deflected less than those with low m/e ratio.
5. Ions with the correct m/z ratio pass through the slit and arrive at the detector.
6. The charge received at the detector is amplified and turned into a sizeable electric current.
7. The electric current operates a pen recorder, which traces a peak on the recording.
8. If the magnetic field is kept constant while the accelerating electric field varies continually, ions of different m/z ratio are deflected one after the other into the detector and a trace is obtained.

Interpreting Mass Spectra
E g. the mass spectrum of rubidium Rb.



Height of peak







m/z 85 87

Note:
1. The height of each peak is proportional to the amount of each isotope present (i.e. it’s relative abundance).

2. The m/z ratio for each peak is found from the accelerating voltage for each peak. Many ions have a +1 charge so that the m/z ratio is numerically equal to mass m of the ion.

Exercise 11
Refer to the diagram of the mass spectrum of rubidium previously to answer this question.

(a) Describe the two isotopes of rubidium using isotopic symbols.
8737Rb; 8537Rb

(b) What information can you get from the heights of the peaks on the mass spectrum?
The height of each peak is proportional to the relative abundance of the isotope it represents. In this case 8537Rb is more than twice as abundant as 8737Rb




Calculating the relative atomic mass of an element

1. Measure the height of each peak.

85 Rb = 5.82 cm
87 Rb = 2.25 cm

Therefore the ratio 85 Rb : 87 Rb is
5.82 : 2.25

2. Calculate the percentage relative abundance
% abundance = amount of isotope x 100
total amount of all isotopes

% 85 Rb = 5.82 x 100 = 72.1 %
(5.82 + 2.25)
% 87 Rb = = 2.25 x 100 = 27.9 %
(5.82 + 2.25)

3. Calculate the Ar
Ar (Rb) = (72.1 x 85) + (27.9 x 87) = 85.6
100

Exercise 12
Use the mass spectrum shown below to calculate:
(a) the percentage of each isotope present in a sample of naturally occurring lithium;
(b) the relative atomic mass of lithium.














3 4 5 6 7 8
mass/charge ratio


Exercise 13
The mass spectrum of neon consists of three lines corresponding to mass/charge ratios of 20, 21 and 22 with relative intensities of 0.910; 0.0026; 0.088 respectively.
Calculate the relative atomic mass of neon. [20.2]


Exercise 14
The percentage abundance of the stable isotopes of chromium are:
5024Cr – 4.31%; 5224Cr – 83.76%; 5324Cr – 9.55%; 5424Cr – 2.38%.

(a) Sketch the mass spectrum that would be obtained from naturally occurring chromium.
(b) Calculate the relative atomic mass of chromium, correct to three significant figures.
(c) Label each peak on the mass spectrum using isotopic symbols.





The next exercise involves a molecular element.

Exercise 15The element chlorine has isotopes of mass number 35 and 37 in the approximate proportion 3:1.


30 40 50 60 70 80
mass/charge ratio

Interpret the mass spectrum of gaseous chlorine shown above indicating the formula (including mass number) and charges of the ions responsible for each peak.



Exercise 16
Calculate the relative atomic mass of potassium, which consists of 93.0% 39K and 7.0% 41K.




Additional Exercises
(a) Chlorine consists of isotopes of relative masses 34.97 and 36.96 with natural abundances of 75.77% and 24.23% respectively.
Calculate the mean relative atomic mass of naturally-occurring chlorine.
35.45
(b) Calculate the relative atomic mass of natural lithium which consists of 7.4% of 6Li (relative atomic mass 6.02) and 92.6% of 7Li (relative atomic mass 7.02). 6.95

(c) Copper (atomic number 29) has two isotopes, the first of relative atomic mass 62.9 and abundance 65%, the second of relative atomic mass 64.9 and abundance 35%.
Calculate the mean relative atomic mass of naturally-occurring copper. 63.6

(d) Using mass spectrometry, the element gallium has been found to consist of 60.4 per cent of an isotope of atomic mass 68.93 and 39.6 per cent of an isotope of atomic mass 70.92.
Calculate, to three significant figures, the relative atomic mass of gallium.
69.7



Mass Spectra of Molecules
Molecules produce more complex mass spectra than atomic spectra. The simplest ion produced is the parent molecule with one electron removed.

M (g) M+ (g) + e-

M+ is referred to as the molecular ion.

It is possible for the molecular ion to break apart to give fragment ions.
M+ (g) Y+ (g) + Z+ (g) etc.
The relative molecular mass of a molecule can be determined from the molecular ion peak.

e.g. The mass spectrum of copper (II) nitrate


molecular ions

peak

height