complete unit 1 notes chem

8/12/2019 Complete Unit 1 Notes Chem

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A Level Chemistry

UNIT 1

THE CORE PRINCIPLES OFCHEMISTRY

NOTES

Written by Mr Sere!nt !n" #r L!$ren%e

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Intr&"'%ti&n

This unit includes the following.

Development of skills in formulae writing, equation writing and calculating

chemical quantities. The study of energetics; definition, measurement and calculation of enthalpy

changes; how the study of enthalpy helps chemists understand chemicalbonding.

The study of atomic structure, introducing a more detailed understanding of

electron configuration and using this to account for the arrangement ofelements in the Periodic Table.

The study of chemical bonding.

Organic chemistry; alkanes and alkenes

Assessment

The nit e!amination will be "hour "# minutes. $t will carry %& marks.

$t will contain two sections, ' and (.

Se%ti&n Ais an ob)ective test *multiple choice questions+.

Se%ti&n (shortanswer and e!tended answer questions.

$t will include questions on analysis and evaluation of practical work.-uality of written communication will also be assessed in this section.



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FORMULAE) E*UATIONS AN# MOLES

Im+&rt!nt terms$n order to use formulae and equations it is necessary to understand the meaning of importantterms.

'tom The smallest unit of an element that can e!ist

olecule The smallest part of an element or a compound which can e!ist aloneunder normal conditions.

$on 'n atom or group of atoms possessing a negative charge

/lement ' substance containing )ust one type of atom.

0ompound ' substance that contains different elements that have been chemicallycombined

/mpirical formulae The simplest ratio showing the different types of atom present in asubstance.

olecular formulae The actual numbers of each type of atom in a molecule of thesubstance.

Am&'nt &, S'bst!n%e

Av!!"r&-s C&nst!nt !n" the M&le$n order to carry out quantitative investigations in chemistry it is necessary to be able to 1count1atoms. This is done using the mole. ' mole contains a given number *the 'vogadro constant+ ofatoms or other particles, and this is measured by mass.

$f we take a mole of atoms of a particular element, the mass will be the 2elative 'tomic ass forthat element.

The 'vogadro constant is the number of particles in " mole of substance *3.&45 ! "&45

+

The m&leis the amount of a substance that contains the same number of particles as there areatoms in "4.&&g of carbon"4. This number of atoms is 3.&4 ! "& 45and is called the Av&!"r&%&nst!nt.

'mount of substance 6 number of particles 'vogadro constant



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M&les ,r&m m!ssTo find the number of moles of atoms we use the formula7

oles 6 mass of substance relative atomic mass

$f we take a mole of ionic lattice or molecules of a material, the mass will be the molar massTo find the number of moles of molecules or compounds we can use the formula7

oles 6 massmolar mass *28, r+

/!amples ". 0alculate the number of moles in "&g magnesium sulphate *g9O:.+ Molar mass of MgSO4 = 24 + 32 + (4x 16) = 120

Moles = 10 /120 = 0.083 mol

4. 0alculate the mass of &.&:mol 0opper*$$+nitrateMolar mass of Cu(NO3)2= 63.5 + (2 x 14) + (6x 16) = 18.5

Mass = Moles x !ormula mass= 0.04 x 18.5 = .5g



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F&rm'l!e "etermin!ti&n

'n em+iri%!l ,&rm'l!is the simplest formula which shows the ratio of each type of atom presentin a compound.

' m&le%'l!r ,&rm'l!is a simple multiple of the empirical formula showing how many of each typeof atom are present in a compound.

The formula of a compound can be calculated from the mass of elements in it.This is done as follows7 9tep "7 8rom the mass of each element find the number of moles. 9tep 47 8rom the number of moles find the simplest ratio. 9tep 57 0onvert the mole ratio to a whole number ratio; this gives the formula.

E.!m+le" #$%ro&ar'o &oss*s of 14.4g &ar'o a% 1.2g #$%roge. Cal&ula*e *s emr&al formula.

C ,

14.4 / 12 1.2 / 1 =1.2 mol =1.2 mol

-a*o = 1 1So *#e emr&al formula s C,

$f we try to work out the structure of this we can see that this is not the formula of the molecule. Tofind this formula we need to know the molecular mass.

$f the molar mass of this compound is %gmol ", what is the molecular formuladm5+ or the amount in moles of a solute present in "dm5 of solution *mol>dm5+.

To make a solution of "mol dm5concentration, "mol of substance is dissolved and the solutionmade up to a total volume of "dm5./!amples1. Cal&ula*e *#e &o&e*ra*o of a solu*o &o*ag 2.4g MgSO4 500&m

3of solu*o.

Moles = 2.4 / 120 = 0.02mol Co&e*ra*o = 0.02mol / 0.5%m3 = 0.04mol%m3

2. Cal&ula*e *#e mass of NaO, reure% *o ma;e 100&m3

of a 0.2 mol %m3

solu*o.

Moles = Co&e*ra*o x


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P!rts +er milli&nThis is a way of e!pressing very dilute concentrations of substances *usually pollutants+.?ust as per cent means out of a hundred, so parts per million or ppm means out of a million.

sually describes the concentration of something in water or soil.One ppm is equivalent to 1 millir!m +er 1////// %m0&, $!ter m2"m03

&r 1 millir!m +er 4il&r!m s&il m2435

or the volume of a gas pollutant in air.One ppm is equivalent to 1 %m0+er 1////// %m0&, !ir

/!ample

" 250 &m3samle of rer a*er as fou% *o &o*a 56 m of gol%.Cal&ula*e *#e mass of gol% *#e 250&m3samle of rer a*er.

9#e samle &o*as 56 m of gol% = 56 mg er %m 3of rer a*er = 56 mg er 1000&m3

1 &m3of a*er &o*as 56 / 1000 mg = 0.056 mg of gol%25 &m3of a*er &o*as 0.056 x 25 = 1.4 mg = 1.4 x 103 g of gol%

/!ampleC#ro& exosure *o CO a* &o&e*ra*os of 0 m or grea*er &auses &ar%a& %amage.oul% $ou 'e a* serous rs; f $ou ere exose% 380 &m 3of &ar'o moox%e a 3500 %m3

room7

380 &m3 3500000 &m3= 380 &m3 3.5 mllo &m3= 380 / 3.5 = 108.6 ppm You are at risk!



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Re!%tin 6'!ntities !n" %hemi%!l e6'!ti&ns

Chemi%!l e6'!ti&ns !n" m!ss %!l%'l!ti&nsole calculations can be used together with chemical equations to determine the quantity ofmaterial formed or used up in a chemical reaction.@hether using masses, volumes of solutions *of known concentration+, or volumes of gases the

same general method can be used.

Ste+ 17 8ind the number of moles of one substanceSte+ 87 se the mole ratio from the chemical equation to find how many moles of the other

substance will be formed > used up.Ste+ 07 8rom the number of moles of this substance calculate the mass > volume >

concentration of the desired substance.

/!ample "7Cal&ula*e *#e mass of ro()sul#a*e (!eSO4) # &a 'e ro%u&e% from .84g of ,2SO4.

8e A B49O: 8e9O: A B4Ste+ 1 Moles of ,2SO4 use% = mass / -!M = .84 / :8 = 0.08

Ste+ 8 9 !rom eua*o 1 mol ,2SO4 forms 1mol of ro()sul#a*eSo mole of ro()sul#a*e = 0.08

Ste+ 0 : Mass of ro()sul#a*e = moles x -!M = 0.08 x 152= 12.2g (3 sg.fg.)

/!ample 4#a* mass of alumum s ee%e% *o rea&* *# .00g of a#$%rous CuSO 47

4'l A 50u9O: 'l4*9O:+5A 50u

Ste+ 1 Moles of CuSO4 use% = mass / -!M = / 15:.5 = 0.043: mol

Ste+ 8 9 !rom eua*o 3 mol of CuSO4rea&* *# 2 mol of "l. So 1 mol of CuSO4rea&*s *# mol of "l.

So 0.043: mol CuSO4rea&*s *# 0.043: = 0.02:3 mol "l

Ste+ 0 : Mass of "l = moles x -!M =0.023: mol 2 g/mol = 0.:0 g (3 sg.fg.)

/!ample 57Cal&ula*e *#e olume &ar'o %ox%e gas # ll 'e ro%u&e% from *#e &om'us*o of 2.20g of

roae (C3,8). 05B% A #O4 50O4 A :B4O

Ste+ 1 Moles of roae use% = mass / -!M =2.20 / 44 = 0.05 mol Ste+ 8 9!rom eua*o 1mol roae forms 3mol &ar'o %ox%e

So mole of &ar'o %ox%e = 0.15 mol Ste+ 0 9


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Titr!ti&ns' titration is a method of finding out how much of one material will react with how much of another

Carryin o!t the titration". ' pipettefiller is added to the volumetric pipette.4. 9ome of the solution is drawn into the pipette. The pipette is tilted and rotated so that all th

surfaces are rinsed in the solution.5. The rinsing solution is then discarded.:. The solution is drawn into the pipette until the bottom of the meniscus is on the mark.#. The solution is then released into a clean conical flask.3. @hen no more solution emerges from the burette, touch the tip of the pipette against the

side of the conical flask. 9ome of the liquid will remain in the tip and this should be left as thepipette is calibrated to allow for this.

. ' suitable indicator should be added to the conical flask%. The flask is placed on a white tile under a burette.C. The flask should be held in the right hand *or writing hand+ and swirled."&. The burette tap should be controlled by the left hand, first and second fingers behind the ta

and the thumb in front of it. 'dd the solution from the burette until the indicator changescolour. ote the reading on the burette. This is the rough reading.

"". Discard the contents of the flask and rinse it with tap water and then distilled water."4. 2epeat the process, adding the solution from the burette fairly slowly with continual stirring.

's the level in the burette approaches that of the rough reading, the solution is added drop bdrop. @hen one drop changes the colour of the indicator, allow the solution to drain down thsides of the burette before taking the reading.

"5. 'ccurate burette readings should be recorded to two decimal places, the second decimalplace being & or #.

":. 2eadings should continue to be taken in this way until one rough and two accurate readingthat are within &."cm5of each other E concordant readings are obtained.

"#. 0alculations should be based on the mean average of the two concordant readings.

Titr!ti&n %!l%'l!ti&nsse the same three step method as before.

/!amplef 23.45 &m3of 0.2 mol %m3so%um #$%rox%e rea&* *# 25.0&m3of sul#ur& a&%.!% *#e &o&e*ra*o of *#e a&%.

4aOB A B49O: a49O: A 4B4O

Ste+ 1 9 Moles of NaO, = 23.45 / 1000 x 0.2 = 0.0046: mol

Ste+ 8 9 Moles of ,2SO4= 0.0046: / 2 = 0.002345 mol

Ste+ 0 9 Co&e*ra*o of ,2SO4 = 0.002345 / 0.025 = 0.0:38 mol %m3 (3 sg.fg.)

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Per%ent!e Yiel"0hemical reactions are carried out in the laboratory and by industry to make new materials. $n theprocess of producing new substances, some material is lost.The efficiency of the conversion process is measured using percentage yield or atom economy.

F Gield 6 'ctual ass of material produced ! "&& Theoretical ma!imum mass which could be produced

/!ample *#e reara*o of &oer sul#a*e a solu*o of sul#ur& a&% &o*ag 2.00g of *#e ure a&%s rea&*e% *# a ex&ess of &oer ox%e.4.3g of *#e #$%ra*e% &oer sul#a*e CuSO4.5,2O as ro%u&e%. Cal&ula*e *#e $el%.

0uO A B49O: 0u9O: A B4O

Moles of sul#ur& a&% use% =mass / -!M = 2.00 / :8 = 0.05 = 0.0204mol

!rom eua*o 1mol sul#ur& a&% forms 1mol &oer sul#a*eSo moles of &oer sul#a*e forme% = 0.0204mol

Molar mass of #$%ra*e% &oer sul#a*e = 63.5 + 32 + 64 + :0 = 24:.59#eore*&al maxmum mass # &oul% 'e ro%u&e% = 24:.5 x 0.0204 = 5.0:g

>el% = "&*ual $el% / *#eore*&al x 100 = 4.3 / 5.0: x 100 = 85.9%

At&m E%&n&my@hereas the percentage yield gives us information about the actual efficiency of a reaction, atomeconomy e!amines the theoretical potential yield of a reaction, by considering the quantity ofstarting atoms in all the reactants end up in the desired product.

Taking the laboratory preparation of copper*$$+ sulphate from sulphuric acid and copper*$$+ o!ide aan e!ample.

0uO A B49O: 0u9O: A B4O ass of starting atoms is ass of desired product is

0uO 6 35.# A "3 6 C.# 0u9O:6 35.# A 54 A 3: 6 "#C.# B49O:6 4 A 54 A 3: 6 C% Total 6 ".#

The atom economy for the production of ethanol from ethene and steam is shown below.

04B: A B4O 04B#OBF atom economy 6 :3 > :3 ! "&& 6 "&&F *there are no unwanted products+.

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"#C.#F atom economy 6 ! "&& 6 %C.CF

".#

olar mass of useful productF atom economy6 ! "&& Total molar mass of starting materials


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#eriv!ti&n &, E6'!ti&ns'n equation can be derived or confirmed by measuring quantities of substances involved in thereaction./quations can be written to e!plain simple test tube reactions such as displacement andprecipitation.

/!ample "1.4g of magaese(


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ENER;ETICS

'll types of chemical reaction involve changes in energy. @hen energy is emitted it can take anumber of forms, but the most common form is as heat energy.Beat energy is called enth!l+y. /nthalpy change is represented by the symbol


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St!n"!r" C&n"iti&nsObviously the quantity of energy given out or taken in during a reaction depends on the amount ofmaterial involved, so the enthalpy values have to be standardised.

The standard enthalpy value 'refers to; molar quantities at 4C%K and " atm pressure *"!"Pa+

Enth!l+y Ch!nesnits of &'are k? molE"7 the mol@1does not refer to one mole of any particular substance, but tomolar quantities in the correct proportions of the equation, and therefore the equation must alwaysbe stated *or implied+7

e.g. &.# 4 A ".# B4 B5 &',6 E:3.4 k? molE"

The enthalpy change here refers to half a mole of nitrogen molecules reacting with ".# moles ofhydrogen molecules to form one mole of ammonia.

$t follows that for; 4 A 5 B4 4 B5

we multiply everything by the same factor and &',6 4 ! E:3.4 6 EC4.: k? molE"

9tate symbols are important. The values for the two reactions below are different.'2 02 '2'2 02 '2*

'llotropic form can also be important;

Craph Cdiam ' 1."mo*-1

9ome important definitions;

(ntha*py chane Definition

/nthalpy of formation, JBLf The enthalpy change that takes place when one mole of

substance is formed from its elements in their standardstate under standard conditions.

e.g. &.# 4 A ".# B4 B5

/nthalpy of combustion, JBLc The enthalpy change that takes place when one mole ofsubstance is completely burned in o!ygen under standardconditions.

e.g. 0%B"% A "4.# O4 % 0O4 A C B4O

/nthalpy of neutralisation, JBLneut The enthalpy change that takes place when one mole ofwater is formed in a neutralisation reaction.

e.g. B0l*aq+A aOB*aq+a0l*aq+A B4O*l+/nthalpy of atomisation, JBLat The enthalpy change that takes place when one mole of

gaseous atoms is formed from the element understandard conditions.

e.g. M0l4*g+40l*g+

N&te (y definition, the1

+' value for an element in its standard state must be Nero.

*$t takes no energy to form an element is its standard state from its constituent element in itsstandard state+.



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E.+eriment!l Me!s'rements &, Enth!l+y/!perimental measurements of enthalpy change are generally based on knowing that the specificheat capacity of water is :.4?g "K". The temperature change of a given mass of water ismeasured, and the energy change is found from7

Beat energy change 6 mass of water ! :.4 ! temperature change *E = m % T3

Or to make sure the sign is always correctBeat energy change 6 mass of water ! :.4 ! *9tarting temp E final temp+

This is converted to an enthalpy change by dividing the energy change by the number of moles *ofthe substance not in e!cess+ and adding the appropriate sign; for e!othermic reactions and A forendothermic reactions.

/!ample#&cm5 of " B0l is placed in a polystyrene cup. #&cm5 of " aOB solution is placed in a beakerThe temperature of each is "%o0. The aOB solution is then added to the B0l in the cup. Thetemperature increases to 4:o0.

0alculate the enthalpy change for this reaction.

Beat energy change 6 "&& ! :.4 ! *"% E 4:+ 6 4#4& ?

oles B0l 6 oles aOB 6 #& > "&&& ! " 6 &. mol

JBL6 4#4& > &. 6 #&,:&& ? mol" or E #&.: k? mol" *5 sig fig+

The two main ways of investigating enthalpy changes are

mi!ing substances in an insulated container, and

using a calorimeter in a combustion e!periment.

' typical e!periment to determine an enthalpy change using an insulated container *to easure thenthalpy of neutralisation of hydrochloric acid and sodium hyrdro!ide+ which takes steps to reduceerrors is given below.

5&cm5is a reasonable volume of liquid to use in an e!panded polystyrene cup.". sing a measuring cylinder, 5&cm5of dilute hydrochloric acid is placed into an e!panded

polystyrene cup.4. ' thermometer is placed in the solution and the temp is recorded every min for 5 mins.5. 5&cm5sodium hydro!ide solution is measured out.:. 't 5.# minutes the sodium hydro!ide solution is added.

#. The mi!ture is stirred continuously.3. 2eadings continue to be taken every minute up to ten minutes.



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' graph of the readings is plotted and the lines e!trapolated as shown in the diagram below.

Qikely errors in these e!periments;o 'ssuming the volume *and therefore mass+ of water used in the calculation is the same as

the volume of the solution used in the reaction.o 'ssuming that the specific heat capacity of the solution is the same as that of water.

' calorimeter could be used to measure enthalpies of combustion. ' typical process for this is outlined below.

". ' measuring cylinder is used to measure out "&&cm5of water.4. The water is placed in a calorimeter *basically a metal can+.5. The liquid to be investigated is placed in a spirit burner. The spirit burner is weighed.:. The temperature of the water is noted#. The apparatus is set up as shown below.3. The liquid is ignited and allowed to heat the water.. 'fter a temperature rise of around 4&o0, the flame is e!tinguished.%. The e!act temperature is notedC. The burner is weighed to find out how much liquid has been burned.

m e t h y * a t e ds p i r i t

w a t e r

Typical results

Qiquid ass of burnerat start >g

ass of burnerat end >g

Temperature atstart >o0

Temperature atend >o0

/thanal, 04B:O ""#.% ""#.#C 4" 4C

Cal&ula*oerg$ = m x C x A = 100 x 4.2 x 8 = 3360BMoles = 0.1: / 44 = 0.0043

, = 3360 / 0.0043 = 813:5B/mol = 81;B/mol (3 sg fg).- 15

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emperat!re

/ oC

ime / mins

3.5 mins

emperat!re

chane +or

reaction


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Hess>s L!$ !n" Enery Cy%les

Bess1s Qaw follows from the "stlaw of thermodynamics *conservation of energy+ and states that;

8or a given chemical change, the overall energy change will always be the same whether thechange takes place in one step or a number of steps. *$t is independent of the route taken+.

@e can use BessRs law to construct cycles, and then equate the enthalpies from differentpathways. 2emember that the direction of the arrow is directly linked to the sign of JB. $f thearrow is reversed, then the sign of JB must be changed *see JB:in the e!ample below+.

) 8

9(

1

32

Consider ) to 8:

c*ocwise antic*ocwise

1 2/

34 /

'

'' ' ' '

'

'

4

E.!m+le 1The energy level of liquid water at 4#o0 will be the same whether it is formed from burninghydrogen or the decomposition of hydrogen pero!ide.9uppose we wanted to find the enthalpy of formation of hydrogen pero!ide, JB Lf*B4O4+.

0hemical equation H83 ? O83 H8O8l3

This reaction does not readily occur, and so we cannot measure its enthalpy directly.Bess1s Qaw can be used to find the value for it using other enthalpy values by constructing a cycle,and then equating the enthalpies from the different routes.

@hen " mole hydrogen is burnt in o!ygen to form water, 4%#.C k? mol "of energy are given out. Thisis the enthalpy of formation of water, JBLf*B4O+.

H83 ? @O83 H8Ol3


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E.!m+le 8 : C!l%'l!tin 1+

' ,r&m enth!l+ies &, %&mb'sti&n

9#e e*#ales of &om'us*o of e*#ee #$%roge a% &ar'o are @140: @286 a%@3:4 ;B mol@1 rese&*el$. !% *#e e*#al$ of forma*o of e*#ee.

*i+ F'll %y%le meth&"0onstruct a triangle with the elements in their standard states *including o!ygen+, ethene A

o!ygen, and the combustion products as the three corners.

sing BessRs Qaw, total enthalpy change from elements to ethene must be the same by either route.

1

+' 6 4

1

+' *0O4+ A 4

1

+' *B4O+ E

1

c' *04B:+

6 4*E5C:+ A 4*E4%3+ E *E":&C+ 6 A :C k? molE"

N.D. "la$s &lu%e *#e + sg


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sing the algebraic method7

1

reaction' A sum of

1

+' *reactants+ 6 sum of

1

+' *products+

4 'l4O5 A 50 :'l*l+ A 50O4

1

reaction' A 4*E"33C+ A & 6 :*A"&+ A 5*E5C:+

1

reaction' 6 E"":4 E *E555%+

6 A 4"C3 k? molE"for 4 mol

so for " mol1

reaction' 6 A "&C% k? molE"

E.!m+le : 'se &, enth!l+ies &, %&mb'sti&n "ire%tlyOccasionally, instead of enthalpies of formation, only enthalpies of combustion *or solution+are given in a question *in /!ample ", two of the enthalpies of combustion were alsoenthalpies of formation+. These cannot be used in e!actly the same way as enthalpies offormation, and you are less likely to make mistakes if you draw the full cycle out.

e.g. 9#e e*#ales of &om'us*o of e*#$e(g) a% 'eEee(l) are @12:: a%

@323 ;B mol@1 a% *#e e*#al$ of aorsa*o of 'eEee s +30;B mol@1

.

!%1

reaction' for 3C2,2(g) C6,6(g) (.e. 3 e*#$e 'eEee)

Meth&"7 construct a cycle, including the combustion products. This means adding .# o!ygenmolecules to both sides.

6 C / 3 ' *2 2

3 C ' / 7.5 2 22 C ' / 7.5 26 6

312"" 3273

'

8rom Bess1 Qaw71

' 6 5 *E"4CC+ E *E545+ 6 E#3: k? molE"



19/73

(&n" Enth!l+y@hen a chemical reaction takes place it involves bond breaking and bond forming.@hen a bond is formed one electron from each atom enters a new electron cloud of lowerenergy. 's the system moves to a lower energy level, energy is given out. $n order to breakthe bond this energy has to be put back into the system.The heat energy obtained when a bond is formed, and taken in when it is broken is called the

bond enthalpy.$f we considered a molecule such as methane, the energy required to break each of thebonds is not e!actly the same. 9o the total energy required to break up a methane moleculeis divided by : to give an !ver!e b&n" enth!l+y.

'verage bond enthalpies can be used to determine !++r&.im!tely the enthalpy changewhich will take place in a particular reaction.

1' re!%ti&n= s'm &, b&n"s br&4en &n le,t : s'm &, b&n"s m!"e &n riht

0onsider the reaction between fluorine and methane.

0B: A :84 08: A :B8

8or each mole(onds broken; 0B : ! :"4 6 "3:% k? (onds formed; 08 : ! :%: 6 "C53 k? 88 : ! "#% 6 354 k? B8 : ! #34 6 44:% k? T&t!l enery 'se"6 A44%& k? T&t!l enery +r&"'%e"6 :"%: k?

Total enthalpy change for the reaction 6 (roken E 8ormed 6 A44%& :"%: 6 "C&: k?.

@e can show this on an energy diagram7

$n reality the reaction between methane and fluorine does not required four fluorine moleculesto collide with a methane molecule at the same instant, but actually takes place as a series ofsteps *in a process called the re!%ti&n me%h!nism+.

(ond energies are useful as a guide to the possible sequence of events.8or e!ample in the reaction between methane and fluorine, which is the first bond to break0B *bond enthalpy :"4+ or 88 *bond enthalpy "#%+onisation

(nery

The energy required to remove one electron from each atom in one mole ofmole of gaseous atoms producing one mole of gaseous ions with onepositive charge.


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$onisation number

To remove the second electron it is necessary to break into a complete shell, so a lot moreenergy needed. 9imilarly with the tenth electron there is an increase in the energy required forits removal as this has to come from a complete shell.

8rom successive ioniNation energies it is possible to tell which group of the Periodic table anelement is in. The table below shows successive ionisation energies for an element./nergy values are given in k?mol".

"st 4nd 5rd :th #th 3th

# "%4& 4:& ""3&& ":%&& "%:&&

otice that there is a large increase after the third ionisation energy E so the fourth electron ismuch more difficult to remove.This means that there are three electrons in the outer shell, so the element is in group 5.

E.!m+le

,ere are su&&esse .. alues ;B mol1

for *o eleme*selement I "&"& "C&& 4C&& #&&& 35&& 810// 4#:&& 4C%&&element G #C& ""#& D/ 3:%& %"4& "&&& "45&& ":3&& "%4&& 4&:&& #"&&. #a* grou of *#e ero%& *a'le are G a% >7

8or I there is a sharp rise between the #th and the 3th values, so I must have five electronsin itRs outer shell. I is a Wroup # element.G shows a )ump between the second and third values, so it is in Wroup 4.

First i&nis!ti&n eneryThe first ionisation energy is energy required to remove one electron from an atom.

The factors which affect ionisation energies are7 the n'%le!r %h!re,

the "ist!n%e !$!y ,r&m the n'%le'sof the electron,

the shiel"in &, the n'%le's by &ther ele%tr&ns.

*(eing spherical s subshells are much better at shielding than p subshells+.

$f first ionisation energy is plotted for a number of elements the results provide evidence forthe e!istence of subshells.



34/73

Qithium has the lowest first ionisation energy of the second period.

$t is more difficult to remove an electron from (eryllium since the charge on the nucleus

has increased (eryllium having a :A nuclear charge, whereas lithium is only 5A.

'lthough (oron also has a higher nuclear charge, it has lower first ionisation energy

because the electron is now in the 1p1 subshell, and the inner [sR subshell shields it fromthe nucleus.

The ionisation energy increases again to 0arbon and itrogen as the nuclear charge

increases.

O!ygen has a higher nuclear charge, but the first ionisation energy is lower because

the e!tra electron now has to pair up in an orbital *in itrogen each p orbital has an

electron in it+ so the e!tra electron is repelled and it is easier to remove it. The values increase after o!ygen as the nuclear charge increases.

The pattern repeats itself for the ne!t period of the periodic table E 8irst ioniNation

energy is a +eri&"i% +r&+erty.



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Peri&"i%ityPeriodicity is the regular occurrence of a trend across a period *or down a group+.any periodic properties can bee!plained in terms of structure and bonding.e.g. /!plaining trends physical properties in period 5 from sodium to argon.

a g 'l 9i P 9 0l 'r

atomU radius > nm &."# &.":& &."43 &."" &.""& &."&: &.&CC E

ion radius > nm &.&C% &.&3# &.: no ion &.4"4 &."%: &."%" E

ion charge A"A5 E5 E4 E" E

structure metal

metal diamond P:\ 9% 0l4 atomic

melting point > K 5" C45 C54 "3%5 5"\ 5C4 "4 %:

boiling point > K ""35 "5C& 44& 435& ##5\ "% 45C %

relative conductivity " ".45 ".C3 semi & & & &

first $./ > k? mol" #&& :& #%& C& "&"& "&&& "43& "#4&

\white P ] red P, a polymer, sublimes at %35

!3 At&mi% r!"ii7The radius of an atom depends on whether it is bonded to another one, and on the nature ofthe bonding *metallic, ionic, covalent+. $f we compare similar types of radius, the general trend

is for atoms to get smaller across a period.This is because the nuclear charge is increasing, and the e!tra electrons being added go intothe same outer shell.On the whole, electrons in the same shell do not shield one another very efficiently, and thesiNe contracts as the effective nuclear charge increases.The alkali metal, in Wroup ", is the largest atom, and the halogen, in group , is the smallest.

b3 I&ni% r!"ii7$onic radii depend on the charge. Positive ions are always smaller than the neutral atoms, andhigher charges make cations smaller still. egative ions are much larger than neutral atoms,and larger still if the charge is higher.8rom the table, it is clear that although a atoms are larger than 0l, a Aions are much

smaller than 0l ions.

%3 Meltin +&ints7These depend on the type of bonding, and the structures of the solids.Wenerally, going across a period metallic bonding at first increases in strength *a^g^'l+ assuccessive atoms have more valence electrons, and then gives way to a giant covalentstructure *9i E like diamond+, and then to molecular solids *P:, 9%, 0l4, 'r+.elting points therefore tend to increase, reaching a ma!imum in the middle, and fall sharplyto a minimum on the right *see table+.



36/73

=alues for the four molecules reflect the siNe of the molecule, and so the intermolecular forces*i.e. 9% _ P: _ 0l4 _'r+.



37/73

"3 (&ilin +&ints7These are less dependent on crystal structure, and so give a better idea of the forcesbetween the particles.

'gain we can note the increase in metallic bonding from a to g to 'l; the high value for thegiant molecule 9i; the remainder are simple molecular structures.8or these, intermolecular forces again depend on siNe of molecule7 P: ^ 9%; then there is a fall

to 0l4and 'r, since they have weaker dispersion forces and lower boiling points.

e3 Ele%tri%!l %&n"'%tivity7'long the series aEgE'l the number of outer shell electrons which can be delocalised inmetallic bonding increases, and the electrical conductivity increases accordingly.(y 9i, however, the effective nuclear charge is too high to allow electron delocalisation, and9i is not a metal, although it is a semiconductor because an electron can be torn freerelatively easily.The remainder are nonmetallic insulators7 their effective nuclear charges are too great toallow electrons to escape under normal conditions.

,3 I&nis!ti&n eneries7The e,,e%tive n'%le!r %h!reis the charge XseenY by an outer electron, after allowing for theeffect of shielding. Thus for sodium *element ""+, which has the configuration e 5s", ten ofthe nuclear charges are effectively shielded by the neon shells, and the effective nuclearcharge on the outer electron is close to A".The first ionisation energies for the first "5 elements are shown below7The general trend !%r&ss ! +eri&"in the Periodic Table, from Qi to e, is for the values toincrease sharply.This is because, while the nuclear charge is increasing by one each time, the additionalelectron is being added in the same outer shell, and generally it is not shielded completely bythe others.Bowever, there are two XkinksY. $n Wroup 5 boron, 4s44p", is lower than e!pected. This isbecause its 4p electron is sufficiently shielded by the 4s4pair for the effective nuclear chargeto be lower than it was in (e.$n Wroup 3 o!ygen, 4s44p:, is also low. This is attributed to the fact that the fourth p electronhas to be spinpaired, and so it is partly repelled by the other electron in the same orbital *alsoof note is that, once an electron is lost, OAhas a halffilled 4p subshell, and in general halffilled and completely filled subshells are particularly stable+.

' similar pattern is followed in Period 5, with low values in Wroup 5 and Wroup 3. ote that ifse&o%ionisation energies are plotted, the pattern is again similar, but this time shifted alongone element e.g. QiA*"s4+ is high, while (eA*"s44s"+ is low.

The general trend "&$n ! r&'+is for ionisation energies to fall by a small amount. 'lthoughthe nuclear charge increases, this is cancelled out by the addition of another closed shell, sothe effective nuclear charge on the outer electron is similar, but the electron is further awayfrom the nucleus, since it is being removed from a higher quantum number shell.



38/73

(ON#IN;

$n the process of bonding electrons move to a lower energy level to form a more stablestructure. This can be done by transferring electron*s+ to another atom or by pairing with anelectron from another atom.

C&v!lent b&n"inTwo non metal atoms can form a stable octet structure by sharing electrons.

any simple covalent compounds obey the &%tet r'le7 the atoms form sufficient covalentbonds to make their outer shells up to eight electrons.

/!ample; chlorine

The shared electrons are positioned between the nuclei of each atom, and there is attractionbetween the positive nucleus of each atom and the shared electrons which holds the atomstogether.

This is called covalent bonding. The two electrons which are shared are called the bondingpair.

ey P&ints &n C&v!lent (&n"in

/lectron deficient atoms share a pair of electrons to form stable electronic structures.

The shared electron pair is attracted to the nuclei of each atom holding the atomstogether strongly.

'toms )oined in this way form molecules.

0ovalent materials may consist of simple molecules, polymers or giant molecules


NUCLEUS NUCLEUS

e

e

C* C*


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8urther e!amples

$f two electron pairs are shared between the same atoms, they form a double bond,as in 0O4, O47

$f three electron pairs are shared between the same atoms, they form a triple bond,as in 47

ote that nonbonding electrons are paired up as Xlone pairsY.

Physi%!l Pr&+erties &, %&v!lently b&n"e" %&m+&'n"s

0ovalent compounds composed of simple molecules have weak intermolecular forcesbetween the molecules. Qittle energy is needed to overcome these forces, so the melting andboiling points of these substances are low.

Wiant molecular substances have strong covalent bonds throughout the structure and sorequire a large amount of energy to overcome them and these substances have high meltingand boiling point.

$n general covalent substances do not have free electrons, so they are nonconductors ofelectricity.Wraphite is e!ceptional as a covalent material in that it does have free electrons and so is aconductor.*Details of diamond and graphite are covered in the X9hapes of olecules and $onsY section+.

- 3" /var/www/apps/conversion/tmp/scratch_6/216168315.doc

' C* C

'

'

' '

D D

Bydrogen chloride

ethane

C


40/73

Ele%tr&n "ensity m!+s/lectron density maps for covalent molecules can be plotted.The electron density map for a molecule such as iodine is shown below.

E.+!nsi&n &, the &%tet'lthough the octet is a particularly stable structure, it is possible for other stable structures toform. 'n atom can increase the number of single electrons it has, and so the number ofbonds it can form by `splitting` an electron pair.

9ulphur is an e!ample.@hat would you e!pect the formula of sulphur fluoride to be< 984

Draw a dot and cross diagram of sulphur fluoride.

$t is possible for an electron pair in sulphur to split. 9plitting an electron pair takes in energy,but the e!tra energy produced by forming e!tra bonds makes it possible.

$f an electron pair in sulphur is split, the sulphur can then form four bonds.

/!ample, 98:

This process splitting electron pairs means that an atom in a compound can have more than %electrons, and it is referred to as e!panding the octet.


E; ;

E E

E; ;

;;

The even distribution of electron

density between the two atoms can beseen in this diagram.


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Other e!amples are 983 and P0l#.

The elements in the second period do not do this because the energy to split the electron pairis too great to be compensated for by the formation of e!tra bonds, but elements belowPeriod 4 form a number of compounds of this type.


E

;

;

;

;

;

;

%C*

C*

C*C*

C*

9ulphur he!afluoride, 983 Phosphorus pentachloride, P0l#


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#!tive %&v!lent b&n"in

C&&r"in!te b&n"inor "!tive %&v!lent b&n"inis similar to normal covalent bonding inthat the bond consists of an electron pair. $t is different in that instead of one electron comingfrom each atom, both the electrons come from a single atom.

0onsider the structure of boron trifluoride and ammonia;

The boron only has three electron pairs in the outer shell. The nitrogen has a lone pair ofelectrons. These two molecules form a dative bond by the nitrogen using the lone pair to

donate to the vacant site on the boron.

The following structure is formed

This can be represented as follows, with the dative bond represented by an arrow;


D'

'

' ;

;

;

D'

'

'

;

;

;

D'

'

'

;

;

;

D'

'

'

;

;

;


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Other e!amples of the dative bond occur in the formation of the hydro!onium and ammoniumions;

H8O ? H? H0O

?

NH0 ? H? NH

?


' '

'

''+ '

+

' '

'

+

D'

'

'

' +

D'

'

'

'

D

'''

'+ +


44/73

Met!lli% b&n"in

etals can be visualised as being made up of positive ions surrounded by a cloud ofelectrons. etals are electropositive; that is they tend to release their valence electrons andform positive ions. $t is these electrons which make up the electron cloud.The structure is held together by the attraction between these positive ions and the electron

cloud or delocalised electrons.

This structure gives metals their properties. 9ince the bonding electrons are not fi!ed theyare said to be "el&%!lise", it means that a small potential difference will cause the electronsto move making metals good electrical conductors./lectron movement also assists in the conduction of heat energy making metals goodconductors of heat. The absence of fi!ed bonds allows the ions to move giving thecharacteristics of malleability and ductility.

F!%t&rs e,,e%tin the strenth &, met!lli% b&n"in7". The more valence electrons the stronger the bonding; as there is a higher

charge on the cations and there is a higher charge in the electron fields.

4. The smaller the metal ion the stronger the bonding; as the electrons becomecloser to the centre of the positive charge.

The metallic bonding determines physical properties of metals. The more valence electronsthere are, the better the conductivity. The stronger the metallic bonding the higher the meltingand boiling points.


+ + + + + + + +

+ + + + + + +

+ + + + + +


45/73

I&ni% b&n"in

etal atoms hold their outer *valence+ electrons weakly. onmetals have a strong pull forthe outer electrons. There is a tendency for metal atoms to transfer electrons to nonmetalatoms.

@hen this happens the metal atom forms a positive ion *so metals are termed electropositive+and the nonmetal atom forms a negative ion *nonmetals are said to be electronegative+. Theions formed have a stable `octet structure`

The strong electrostatic attraction between the cation and the anion lead to a strong ionicbond. This forms a giant structure built up of these oppositely charged ions.

The strong electrostatic forces within the giant structure mean that it requires a large amount

of energy to break them down. 's a result of this, ionic compounds have high melting andboiling points.

The ionic compounds are unable to conduct electricity in solid form, but when dissolved inwater or in molten from the ions are free to move and so they conduct.


NUCLEUS NUCLEUS

9odium chloride is an ioniccompound. The structure of thiscompound is shown here.


46/73

ey P&ints &n I&ni% (&n"in

'n electron *or electrons+ is transferred form one atom to another to form stable

electronic structures.

$n this process positive and negative ions are formed

The oppositely charged ions attract have very strong electrostatic forces ' giant structure of ions is formed

The compounds have high melting and boiling points

The compounds do not conduct as solids, but do when molten or in solution.

Re+resentin i&ni% %&m+&'n"s$onic substances are giant structures, but the essential electronic structure of an ioniccompound can be represented by showing one set of ions.

/!ample 9odium chloride

9imple electronic configuration 9odium7 4, %, " 0hlorine7 4, %,

To form stable structures an electron is transferred from sodium to chlorine.

9odium ion7 4, % 0hloride ion7 4, %, %

' diagram to the ionic compound shows )ust the outer shell.

/!ample 4 0alcium bromide 0alcium7 4, %, %, 4 (romine7 4, %, "%,

To form stable structures two electrons are transferred from calciumone each to two bromine atoms.

0alcium ion7 4, %, % (romide ion7 4, %, "%, %


Da

+

C*

-

Ca2

r-

r-


47/73

Ele%tr&n "ensity m!+sI rays can be deflected off solids to give us information about their structure. I ray diffractionof ionic solids can be presented as an electron density map. 'n electron density map showslines which represent areas of equal electron density.

$n practice the situation is more comple! as the cation can distort the electron field of theanion.

Sies &, i&ns in ! ;r&'+Woing down a group of the Periodic Table, the ions formed have the same charge.

's the atoms from which they were formed become larger so do the ions.

Sies &, is&ele%tr&ni% i&ns$soelectronic ions are ions that have the same number of electrons.

'tom O 8 a g 'l

umber ofelectons

% C "" "4 "5

8ormsions by

Waining 5electrons

Waining 4electrons

Waining "electron

Qosing "electron

Qosing 4electrons

Qosing 5electrons

8ormula ofion

5 O4 8 aA g4A 'l5A

umber ofelectrons

"& "& "& "& "& "&

9iNe of ion

>nm

&."" &.":& &."53 &.&C# &.&3# &.&

's the ions all have to same number of electrons, the siNe of the ion is determined by thecharge on the nucleus. 's the nuclear charge increases from to 'l so does the attraction ofthe electrons to the nucleus and so the ionic siNe decreases.


'n idealised electron map for an ion pair

in an ionic compound is shown to theright. $t is important to remember that thelines do not represent shells.

The electron density map for sodiumchloride is shown to the left.The distortion of the ions, particularly thechloride can be seen.


48/73

I&ni% %&m+&'n"s !n" Enery

/nergy changes take place when ionic compounds are formed.The process of bond breaking is endothermic and the process of bond breaking ise!othermic. The energy changes that take place when an ionic compound forms are shownbelow.

(&n" bre!4in

/nergy taking in to convert the metal lattice into individual metal atoms.

/nergy taking in to convert the nonmetal molecules to individual nonmetal atoms.

I&n ,&rm!ti&n

/nergy taking in to convert the individual metal atoms into individual metal ions.

/nergy given out when the individual nonmetal atoms gain an electron to from an

anion.

(&n" ,&rmin

/nergy given out E as the cations and anions come together to form an ionic lattice

the L!tti%e Enth!l+y.

This process is shown below for sodium chloride.

9uch an energy diagram is called a (&rn9H!ber %y%le.*$t is not necessary to construct a (ornBaber cycle for nit "+.


Sodium chloride

Sodium metal + chlorine molecules

Eodi!m atoms ch*orine atoms

Eodi!m ions ch*orine atoms

Energy to atomise sodium

and chlorine

Energy to ionise sodium

Energy given as compound

forms

Eodi!m ions ch*oride ions

Energy given as chloride

ions form

Energy given as the

sodium chloride lattice

forms


49/73

The Qattice /nthalpy can be calculated from this.

JBQlat6 JBfE *JBatSa A JBatS0l A $/Sa+ E /'*0l+

8or sodium chlorideJBQlat6 :"" E *"&C A "4" A :C:+ E 53: 6 "k?mol

"

The values obtained from these (ornBaber cycles have to be found by e!perimentalmeasurement of the various energy values and so are often referred to as e.+eriment!lly"etermine" l!tti%e enth!l+ies5They represent the actual lattice enthalpy, as opposed to a theoretical lattice enthalpy, whichis based on a theoretical model for ionic compounds.

0onsidering (ornBaber cycles allows us to see why compounds have the formulae they do.

Take calcium chloride as an e!ample.2elevant values

The enthalpy of formation can be found by rearranging the equation used above.JBQlat6 JBfE *JBatS0a A JBatS0l A $/S0a+ E /'*0l+JBf6 JBQlatA *JBatS0a A JBatS0l A $/S0a+ A /'*0l+

8or 0a0l JBf6 4& A "C5 A "4" A #C& 53: 6 "%& k?mol"

8or 0a0l4 JBf6 445 A "C5 A "4" A #C& A ""#& A 4! 53: 6 C& k?mol"

ore energy is needed to form 0a4Abut this is outweighed by the much larger lattice enthalpy,so the formation 0a0l4of is energetically more favourable than for 0a0l.

The lattice enthalpy of 0a0l5is greater than for 0a0l4, but the e!tremely large amount ofenergy needed to form the 0a5Amakes the process energetically unfavourable.

F!%t&rs in,l'en%in l!tti%e enth!l+y5- 4"

/var/www/apps/conversion/tmp/scratch_6/216168315.doc

"stioniNation energy of calcium #C&k?mol"

4nd

ioniNation energy of calcium ""#& k?mol"

5rdioniNation energy of calcium :C:& k?mol"

/lectron affinity of chlorine 53: k?mol"

Qattice enthalpy of 0a0l 4& k?mol"

Qattice enthalpy of 0a0l4 445 k?mol"

'tomiNation enthalpy of 0a "C5 k?mol"

'tomiNation enthalpy of 0l "4" k?mol"


50/73

The lattice enthalpy of an ionic compound is affected by the siNe and charge on the ionswhich make it up.

' decrease in the siNe of any ion increases the lattice enthalpy. This is because small ionscan be close together and the smaller distance of separation the larger the attractive forcebetween the ions.

'n increase in charge also increases lattice enthalpy. This is because the force of attraction

between ions increases as the charge on the ion increases.



51/73

The&reti%!l "etermin!ti&n &, l!tti%e enth!l+y$t is possible to calculate the lattice enthalpy by visualiNing the ionic lattice as made up ofperfectly spherical ions in a lattice. The lattice enthalpy is determined theoretically bycalculating all the forces of attraction and repulsion in the lattice.

' simplified version of this is given below7

BQ/o 6 E constant V

+

+

+

rr

zz i.e. +r&"'%t &, %h!res s'm &, r!"iiJ

Bere, the constant depends on the type of structure *the way in which the ions pack together+.NAand NEare the charge on the cation and anion respectively, and rAand rEare the radii ofcation and anion.

' comparison of the lattice energies of a0l *E& k? molE " + and g0l4*E4#&& k? molE " +

shows a much more negative value for the magnesium compound7 not only is NA double forg4A, but also its ion has a smaller radius *rA+, and this too leads to a larger lattice energy.

C&m+!rin e.+eriment!l !n" the&reti%!l v!l'es ,&r l!tti%e enth!l+yThe lattice energy is the energy change which occurs when well separated gaseous ions arebrought together to form a crystal. 's stated above, it is possible to obtain a theoretical valueof the lattice energy by considering all the attractions and repulsions between the ions in thelattice.

The values obtained by using (ornBaber cycles are determined by e!periment and reflectthe actual value.

The table below compares theoretical lattice energies with e!perimental lattice energies.

0ompound /!perimental value> k?mol" Theoretical value> k?mol" Difference invalues > k?mol"

g84 4C# 4C"5 ::

g$4 454 "C:: 5%5

The compound magnesium fluoride has theoretical values which agree quite well with thee!perimental value, but the magnesium iodide values agree less well. $n general the iodidestend to show a greater difference between these values than fluorides.

The e!perimental values show that lithium halides are actually more stable than would bee!pected by regarding them as purely ionic compounds. $n the same way, iodides are in

reality more stable than would be e!pected by treating them theoretically as purely ioniccompounds. This discrepancy is caused by a covalent character which e!ists in the ioniclattice.The greater the difference between the e!perimental value and the theoretical value, thegreater the degree of covalency in the lattice.This covalency is caused by +&l!ris!ti&n.



52/73

P&l!ris!ti&n in I&ni% (&n"in@hen an ionic bond is formed, the metal atom transfers one or more electrons to the nonmetal and a cation *A+ and an anion *+ are formed.

$n perfect ionic bonding there is complete separation of charge. $n polarised ionic bondingsome of the electron charge on the anion is pulled back towards the cation.

Polarisation occurs to different e!tents according to the ions involved

' highly ionic material

Polarisation

0onsiderable polarisation

The e!tent of the polarisation depends on how much the cation pulls on the electrons; the+&l!risin +&$er, and on how much the anion allows the electrons to be pulled away7 the+&l!ris!bility.

Polarising power is increased by higher positive charge and smaller cation.

Polarisability is increased by higher negative charge and larger anion.


Cation

+

)nion

-

The cation attractselectrons in the anion

Cation

+

)nion

-

The electron field aroundthe anion is distorted


53/73

F!%t&rs $hi%h in%re!se +&l!ris!ti&n in i&ni% %&m+&'n"s ". increased charge on anion or cation, 4. increased siNe of anion, 5. decreased siNe of cation.

Qow polarisation Bigh polarisation

low charge *A","+ high charge *A5,5+ small anion large anion large cation small cation

e.g. KA8 e.g. 'l5A *$+5

$t is of course incorrect to regard ion pairs in isolation, because they actually e!ist in a lattice.

Polarisation of ionic compounds gives the compounds a covalent character.The greater the polarisarion, the greater the covalent character.The e!tent of this polarisation can be found by comparing theoretical and e!perimental lattice

enthalpies.

@hen such a comparison was carried out with selected compounds in the previous section, itwas seen that lithium and iodide compounds demonstrate significant covalent character.

Qithium has a high polarising power and iodides have a high polarisability.The more polarised the compound, the greater the difference between the theoretical valuesand the e!perimental values.

8urther e!ample

compoundQattice enthalpy value based on theoretical

ionic model

Qattice enthalpy value based on

e!perimentsodiumchloride

33." k?mol" 3.: k?mol"

silver chloride 3%.3 k?mol" C"3.5 k?mol"

9odium chloride is almost "&&F ionic, hence the good agreement.9ilver chloride has a poor agreement and so has some covalent nature in its bonding.



54/73

INTRO#UCTORY OR;ANIC CHEMISTRY

Intr&"'%ti&nOrganic chemistry is the study of carbon compounds e!cluding carbon mono!ide, carbondio!ide and carbonates. 0arbon atoms usually form four bonds. They have the unusual

property of being able to )oin with other carbon atoms to form chains. This is known as%!ten!ti&n.

Organic chemicals occur in series such as alkanes, alkenes, alcohols and halogenoalkanes.These series are known as h&m&l&&'s series.

'll the compounds in a particular homologous series have the same general formula andsimilar chemical properties.

N&men%l!t'reThere are two components to the name of an organic compound.

The first part of the name is given based on the number of carbon atoms in the molecule.

First +!rtof name

umber of0 atoms

ame

" Meth9

4 Eth9

5 Pr&+9

: ('t9

# Pent9

3 He.9

He+t9% O%t9

C N&n9

"& #e%9

The second part of the name is given according to which homologous series they belong.'lkanes always have name endings of 9ANE.The part of the molecule that determines which group it is a member of, is called the,'n%ti&n!l r&'+.



55/73

Se%&n" +!rtof name

8unctional group ame of group ame ending

'lkene 9ene

0hloro Chl&r&9

Pl!%e" !t ,r&ntbromo (r&m&9Pl!%e" !t ,r&nt

$odo I&"&9Pl!%e" !t ,r&nt

'lcohol 9&lSee n&te bel&$

'ldehyde 9!l

Ketone 9&ne

0arbo!ylic acid 9&i% !%i"

$f a molecule contains an OB group, it will end with the name Eol.

$f it contains another group, such as a carbo!ylic acid, the alcohol is named as hydroy infront on the name, and the oic acid at the end.


C*$

C C

r$

>$

'$

C

'

$

C

$

$

C

$'


56/73

R'les ,&r n!min sim+le &r!ni% %&m+&'n"s

*i+ $dentify the longest XstraightY carbonEchain *backbone+.

*ii+ umber the carbonchain from the end with the most functional groups attached. *9omefunctional groups, like E0OOB, must be the first carbon atom, if possible+.

*iii+ Qist the functional groups in alphabetical order, each prefi!ed by the number*s+ of thecarbon atom in the backbone to which it is attached. Dashes separate numbers andwords.

C' C' C' C'C'3 2 2 3

C'3

2-methy*pentane

C' C C' C'

rC*

2

C'3

2 3

2-@romo-1-ch*oro-2-methy*@!tane

*iv+ 9ome groups appear at the e% of a name. umbering here starts from the end of thechain which is nearest to one of these groups7

e.g. 0B50B(r0B*OB+0B5 is 5bromobutan4ol

E.!m+les

propan4ol 0B50B*OB+0B5

the dash separates numbers and words.

4,#dimethylheptane 04B# 0B5

0B50B0B40B40B40B5

"bromo4chloroethane 0B40l0B4(r

(ut"ene 0B5 0B4 0B6 0B4

",4Dibromopropane 0B50B*(r+0B4(rnumbers are separated by commas



57/73

Ty+es &, ,&rm'l!e

The em+iri%!l ,&rm'l!of a substance shows the simplest ratio of atoms in the compound*e.g. for alkenes, 0B4+, while the m&le%'l!r ,&rm'l!shows the actual numbers of atoms ofeach element in one molecule *e.g. 0:B%for butene+.

The semi9str'%t'r!l ,&rm'l!of a compound shows the order in which the atoms are bondedtogether in one molecule of the compound7 from it one must be able to work out all the atomsto which each particular atom is bonded *e.g. 0B50B40B60B4for but"ene; 0B50B60B0B5for but4ene+.

The str'%t'r!l ,&rm'l!shows both the relative placing of the atoms, and the number ofbonds between them7

' C C

'

'

'

ethanoic acid

C

C C

C

'

'

' '

'

''

cis@!t2ene

'

The s4elet!l ,&rm'l!is a simplified version showing bonds between 0 atoms and anyfunctional groups, but omitting other 0 and B atoms. /ach end of each line in the mainskeleton shows a 0 atom, but a bond to a functional group does not end in a 0 *e.g. 0](rbecomes ](r, and 0]OB becomes ]OB, while 0]0B5becomes ] +. /!amples7

butan4ol :bromo4methylpent4ene

' ener!l ,&rm'l!is used to represent members of a homologous series.e.g. al;aes0nB4nA4 or 0B5*0B4+!0B5

rmar$ al&o#ols0nB4nA"OB or 0B5*0B4+!OB./ach member of a homologous series differs from the ne!t by one CH8group7


'

r


58/73

Usin &, Or!ni% Chemi%!ls

any organic chemicals and reagents used in their reactions have properties that mean theyneed to be handled with care.The dangers presented by these chemicals such as flammability, being irritant or beingcarcinogenic, are called haNards.

@hen these materials are handled, the likelihood of the haNards they present being realiNed iscalled the ris4.$n a practical using haNardous chemicals, the likely dangers and how they can be minimiNedis carried out in a ris4 !ssessment.

@ays in which risk can be minimiNed include;

@orking on a smaller scale

Taking specific precautions or using alternative techniques

se an alternative procedure that involves less haNardous substances



59/73

Al4!nes !n" Al4enes

The Al4!nes

'lkanes are hy"r&%!rb&ns*i.e. compounds of carbon and hydrogen only+.They are called s!t'r!te"hydrocarbons because they contain no double bonds, and so

cannot undergo addition reactions.

They have the general formula 7 0nB4nA4

Physi%!l Pr&+ertiesThe boiling points of the alkanes increase steadily with the number of 0 atoms.This is because the van der @aalsR forces between the molecules increase with the totalnumber of electrons in the molecule. 's the chain length increases, the van der @aalsRattractions increase, and the liquid needs to reach a higher temperature before the moleculescan acquire enough energy to separate and go into the vapour state.

's isomers become more highly branched, boiling points usually fall slightly.

This is because the branching means that the molecule has less surface area in contact withits neighbours, thus decreasing the van der @aals attractions.

Is&merism

There are two main types of isomerism7 str'%t'r!l is&merismand stere&is&merism.

Str'%t'r!l is&mers7 these have different structural formulae. i.e. the atoms are )oinedtogether in a different order.

't the most e!treme, they may have different functional groups7 e.g. 04B3O may be ethanol,0B50B4OB, or it may be metho!ymethane *an ether+, 0B5O0B5.

'lternatively, they may have similar functional groups, but differ in the position of these in themolecule7 e.g. propan"ol, 0B50B40B4OB, and propan4ol, 0B50B*OB+0B5.Or they may differ only in the hydrocarbon chain *see below+7

B50 0B4 0B 0B4 OB

0B5

B50 0B 0B4 0B4 OB

0B5

2methy*@!tan1o* 3methy*@!tan1o*

Stere&is&mers5These have the same structural formulae, and differ only by the arrangement of the atomsin space. There are two sorts7 cistrans *or geometrical+, and optical. Optical isomerism willbe looked at ne!t year.

- 5" /var/www/apps/conversion/tmp/scratch_6/216168315.doc

$somerism is when two molecules of the same molecular formula have a differentarrangement of atoms *structural formula+.


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Is&mers &, !l4!nes'lkanes e!hibit structural isomerism.

e.g. $somers of pentane

B5

0 0B4

0B4

0B4

0B5

B5

0 0B 0B4

0B5

0B5

Pentane methylbutane

B50 0 0B5

0B5

0B5

Dimethylpropane

e.g. $somers of he!ane

B50 0B4 0B4 0B4 0B4 B50 0B 0B4 0B4

0B5

0B5 0B5

Be!ane 4methylpentane

B50 0 0B4

0B5

0B5

0B5 B50 0B 0B

0B5

0B5 B50 0B4 0 0B4

0B5

0B5

0B5

4,4dimethylpropane 4,5dimethylpropane 5methylpentane



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F'els

One of the main uses of alkanes is as fuels.0rude oil consists of a mi!ture of a large number of hydrocarbon compounds.0rude oil undergoes ,r!%ti&n!l "istill!ti&n, and the fractions obtained by this process consistof mi!tures of hydrocarbons that boil within particular ranges.

The more useful fractions are those with the lower boiling ranges, such as that of gasolinewhich is used for petrol, but they generally occur in smaller proportions.

To convert some of the larger, less useful alkanes into smaller, more useful alkanes theprocess of %r!%4inis used.0racking involves passing the alkane vapour through a heated catalyst.The alkane molecules break up to form a smaller alkane molecule and at the same time analkene, such as ethene, is formed which can then be used to make polymers.

' typical cracking reaction might be7

0"&B44 0%B"% A 04B:

The straight chain alkanes do not burn very evenly, and tend to cause X4n&%4inY in a carengine. (ranched alkanes burn more smoothly and are much more appropriate for modernhighperformance engines.The petrochemical industry carries out the process of re,&rmin to convert straight chainalkanes into branched chain alkanes.

Bydrocarbon fuels produce carbon dio!ide when they burn.0arbon dio!ide is a Xreenh&'se !sY. @hen infra red radiation from the sun hits the /arthRssurface, it is absorbed and then reradiated at a lower frequency.

' greenhouse gas absorbs the reradiated infra red and converts it to heat energy, so causinga warming of the atmosphere. This means that the production of carbon dio!ide by burning

fossil fuels increases the concentration of greenhouse gases in the atmosphere and isconsidered to cause %lim!te %h!ne.The problem of climate change means that there is a search for alternative fuels which do notplace greenhouse gases in the air.



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Chemi%!l +r&+erties &, !l4!nes

'lkanes contain carbonhydrogen bonds, and, from ethane on, carboncarbon single bonds.

These are all nonpolar, and the electron density in the bonds is hard to distort *not

polarisable+, so that alkanes do not react with electrophiles or nucleophiles, or other polarreagents, like acids, alkalis or aqueous o!idising agents.

The bonds can only undergo h&m&lyti% b&n" ,issi&n, whichoccurs when a covalent bondbreaks in such a way that each atom takes one electron. This leads to formation of two ,reer!"i%!ls5

' ,ree r!"i%!lis an atom, or group of atoms, which have an unpaired electron. $t is normallyhighly reactive, and written with a dot *which does not show an e!tra electron, simply theunpaired one+. e.g. 0l or 0B5.

'lkanes therefore undergo very few reactions7 %&mb'sti&nand %r!%4inare the mostimportant.

Re!%ti&ns &, !l4!nes

!3 C&mb'sti&n'lkanes burn in a plentiful supply of air or o!ygen, to give carbon dio!ide and steam7

CH ? 8O8 CO8 ? 8H8O

$f air is restricted, carbon mono!ide is likely to be formed7

CH ? 1@ O8 CO ? 8H8O

The former reaction gives out much heat, and so alkanes are useful fuels. ethane itself iscalled natural gas, and is used in the home and in modern gasturbine power stations.

b3 H!l&en!ti&n$t is possible for the hydrogen atoms in the alkane to be s'bstit'te" by chlorine or bromine.

'lkanes react with chlorine or bromine, in the presence of light, or on heating, to formhalogenoalkanes and the hydrogen halide7

8or e!ample the reaction between methane and chlorine can be represented as follows7


' '

'

'

C C*C*+ ' C*'

'

'

C C* +


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The reaction does not necessarily stop at one substitution, and the reaction between methaneand chlorine produces dichloro, trichloro and tetrachloro methane.

0B: A 0l4 0B50l A B0l

0B50l A 0l4 0B40l4A B0l

0B40l4 A 0l4 0B0l5A B0l

0B0l5A 0l4 00l:A B0l

The process of substitution is random, so all possible products result, though if there ise!cess methane the main one will be 0B50l, while a large e!cess of chlorine will give mainly00l:.

The substitution of a hydrogen atom by a chlorine atom actually takes place in a number ofstages. Qooking in detail at what happens in each stage of such a reaction is called the

re!%ti&n me%h!nism.The halogenation of an alkane takes place by a ,ree r!"i%!lreaction.

The reaction between alkanes and these halogens requires an energy source such as ultraviolet *u.v.+ light. 0onsidering the reaction between methane and chlorine, looking at thebond enthalpies for the bonds present

(ond /nergy k?>mol

0B :"5

0l0l 4:4

$t can be seen that the 0l0l bond requires less energy to break, so this is the bond which isbroken by the u.v. light.

This energy is used to split the halogen molecule to form two chlorine atoms; the chlorineatoms have an unpaired electron E such a species is called ,ree r!"i%!l.

The chlorine radical reacts with a methane molecule. $t forms a bond.Qooking at the bond enthalpies for the bonds that could be formed;

(ond /nergy k?>mol

00l 55C

B0l :5"

$t can be seen that the B0l bond has a higher bond enthalpy, so this is the bond which forms.


C*C* C*C*


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This forms a methyl radical. The methyl radical then reacts with a chlorine molecule7

$n this reaction a chlorine radical is generated; this then reacts with a methane molecule andthe process continues.


C

'

'

' 'C*+ 'C

'

'

' +C*

C*C*C

'

'

' +

+ C*C

'

'

' C*


65/73

C'rly !rr&$ me%h!nisms

2eaction mechanisms look at the movement of electrons within and between molecules andions. The movement of electrons are shown by Xcurly arrowsY, using for a singleelectron and for an electron pair.

The first step in the reaction between methane and chlorine is the formation of the chlorineradical. This is called the initi!ti&n ste+. $t is shown as follows.

The chlorine radical then attacks the methane molecule.

0B: A 0l 0B5 A B0l

The methyl radical from this reaction reacts with a chlorine molecule to form chloromethaneand a new chlorine radical.

0B5 A 0l4 0B50l A 0l

The chlorine radical starts the cycle again and the process continues.These two reactions are called the +r&+!!ti&n st!e.

The propagation stage continues until two radical meet to form a molecule.There are three possibilities for this reaction. 9ince the ends the sequence it is called thetermin!ti&n st!e.

0l A 0l 0l4

0l A 0B5 0B50l

0B5 A 0B5 04B3

The overall mechanism is called ,ree r!"i%!l s'bstit'ti&n.

ote that in the propagation stage the 0l at the end can go on to attack another 0B:and sothe chain can go on for several thousand reactions from one initiation.

The presence of traces of 04B3in the products shows that 0B5 radicals must have beenformed.

Of course, the 0l atoms can remove a hydrogen atom from 0B50l, and so the reaction cango on to 0B40l4, then 0B0l5and finally 00l:.


C*C* C*C*


66/73

The Al4enes

'lkenes are hydrocarbons that contain one double bond.

They are 'ns!t'r!te" %&m+&'n"s7 compounds which are able to undergo additionreactions.

They have the general formula 7 0nB4n

The double bond in alkenes does not consist of two identical bonds.One of the bonds is a normal bond with the electron cloud lying between to two atoms. Thisis called a sigma, , bond.The other bond has an electron cloud which is placed above and below the two atoms. Thisis called the pi, , bond.

The electron clouds in ethene can be represented as shown below.

Is&mers &, !l4enes */ $somerism+

The geometry of the 060 bond is determined by the bond, and means that ethene is arectangular, flat molecule.

0 0

B

B B

B

The bond angles are at about "4&o, and rotation about the 060 a!is is restricted, sincetwisting through C&owould mean the porbitals would no longer overlap and the bond wouldbe broken.

The restriction of rotation produces isomers called E9 is&mers*geometric isomers or cistrans isomers+.



67/73

8or / isomerism *cistrans isomerism+ to occur there must be a "&'ble b&n"and t$&"i,,erent r&'+son e!%hof the double bonded %!rb&natoms.

0

0

B2

B 2

0

0

2B

B 2

.

/ and cistrans are different terms for the same type of isomerism.

Trans refers to the isomer where the 2 groups are on opposite sides of the 060.0is refers to the isomer where the 2 groups are on the same side of the 060.

Is&mers &, b't!ne

0 0

B

B50 B

0B5

0 0

B

B50 0B5

B

0 0

B50

B50 B

B

4/ but4ene 4 but4ene 4Emethylpropene trans but4ene cis but4ene

The / *cis > trans+ isomers will be chemically very similar, though not quite identical, and willhave slightly different melting and boiling points.ote that a third isomer, 4Emethylpropene, which has two methyl groups attached to thesame carbon atom, is a str'%t'r!l isomer of these, as is but"ene

E9 !n" 9 vs %is9tr!ns n&men%l!t'reThe cis trans system breaks down with e!amples like;

8

0l

8

0l

The / naming system is more useful than the cistrans system.

The naming of the isomer is determined by priorities.

/ach atom on the double bond is given a priority determined by its atomic number.

The smaller the atomic number the lower the priority E hydrogen in the e!ample above.

$f atoms with the lowest priority are on the same side on each end of the 060 the

isomer is called the 9is&mer. *6usammen *together++.

$f atoms with the lowest priority are on the same side on each end of the 060 the

isomer is called the E9is&mer. */6/ntgegen *opposite++.


/ isomer *trans+ isomer *cis+


68/73

Re!%ti&ns &, !l4enesThe alkenes are more reactive than the alkanes because of the 060 bond.$t is possible for the double bond to break, allowing each carbon to form a new bond, which isoften energetically favourable. Balogens, hydrogen halides, hydrogen and potassiummanganate*=$$+ will produce !""iti&n re!%ti&nswith alkenes.

A""iti&n &, hy"r&en hy"r&en!ti&n3'lkenes react with hydrogen in the presence of a nickel catalyst at"#&o0.

This is important in the hydrogenation of vegetable fats to make margarine.'n unsaturated vegetable oil is mi!ed with a nickel catalyst at #&o0,and hydrogen gas

bubbled through. The saturated product has a higher melting point, and is less easily o!idised*longer shelf life+.

A""iti&n &, br&mine'lkenes will decolorise bromine immediately7

The decolorisation of bromine water is used as a test for the 060 double bond.This reaction occurs in an inert solvent like volasil.

A""iti&n &, %hl&rine


+ ''

'

$

'

CC $ $

''

CC $

''

+ C*C*

'

$

'

CC $ $

''

CC $

C*C*

'

$

'

CC $ + rr $

''

CC $

rr


69/73


70/73

A""iti&n Re!%ti&ns &, Al4enes

The addition reactions have a mechanism which we need to study.$t is the electron rich area in the double bond which causes the initial attack.

' molecule attacking such an area is called an ele%tr&+hile*electron loving+. This reactionresults in addition, so this type of reaction is called ele%tr&+hili% !""iti&n.

' definition of the term electrophile is;'n electrophile is a species attracted to an area of negative charge and is an electron pairacceptor.

Me%h!nism ,&r the re!%ti&n &, hy"r&en br&mi"e $ith eth!ne5Qooking at dot and cross diagram for the reaction.This is n&t the $!y ! me%h!nism is sh&$n5


'

' '

'

C C

'

r

". The hydrogen in the hydrogen bromidehas a partial positive charge. $t is attracted tothe high electron density in the double bond.

4. The hydrogen draws an electron pair outof the double bond and forms a bond to thehydrogen.

'

' '

'

C C

'

r

5. The bond between the bromine and thehydrogen weakens and breaks, both theelectrons going to the bromine.

'

' '

'

C C'

r -

+

:. The carbon on the right now has onlythree of its original four electrons, so it hasa positive charge. This intermediate iscalled a %!rb&%!ti&n i&n.

#. The bromine has gained an electron

making it a bromide ion with a negativecharge.

3. The bromide donates an electron pair tothe positively charged carbon forming a bond.

'

' '

'

C C' r


71/73

The reaction mechanism for this is shown as follows.

The mechanism for the attack of bromine on ethane is almost identical, but as bromine is nota polar molecule it has no partial charge.

's a bromine molecule approaches the ethane, the high electron density in the double bondrepels electrons in the nearest bromine so causing an induced partial positive charge.


'

' '

'

C C

'

r

'

' '

'

C C'

r

+

-'

' '

'

C C' r

'

' '

'

C C

r

'

' '

'

C C

r

+

-'

' '

'

C C r

r

r r


72/73

Asymmetri% A""iti&ns M!r4&vni4&,,>s r'le3

The addition of hydrogen bromide to propene could lead to two possible intermediates.

The carbocation ion with the charge on the central carbon is formed as the charge is

stabilised by the two carbons on each side of it. 9o the mechanism is as follows

M!r4&vni4&,,>s r'lestates that;@hen addition of a hydrogen halide takes place, if the alkene is not symmetrical, thehydrogen adds to the carbon that already has the most hydrogen.


'

' '

'

C C

'

r

'

'

C '

' '

'

C C

'

'

C '

' '

'

C C

'

'

C

'

'++

$f the hydrogen )oins the central carbon the ion on the left will be formed.$f the hydrogen )oins the end carbon the ion on the right will be formed.

'

' '

'

C C

'

r

'

'

C '

' '

'

C C

'

'

C '

+

r-

'

' '

'

C C

'

'

C '

r


73/73

Test ,&r !l4enesThe addition reactions are used to test for the presence of a double bond.$f br&mine $!teris added to an unsaturated compound, an addition reaction occurs fairlyrapidly and the bromine water changes from orange to colourless.$n this test the product contains a (r and an OB.This is because the bromine water contains hydro!ide ions from water.

The reaction mechanism for the test on ethene shows why this product is formed.

A""iti&n P&lymers'n addition polymer is formed when monomers containing double bonds are polymerised.$n the process of polymerisation the double bond breaks and the molecules )oin to formmolecules of thousands or millions of units.

Polymer structure of structure ofonomer polymer

/thene

Propene

0hloroethene

'

' '

'

C C

r

'

' '

'

C C

'

+

-'

' '

'

C C '

r

r r

OB from the water takes placein the second step.

C C

' '

' 'C C

' '

' ' n

C C

' C'3

' 'C C

' C'3

' ' n

; ;

C C

' C*

' '

C C

' C*

' '

C C

' C*

' ' n

complete unit 1 notes chem

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