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1© Pearson Education Ltd 2008This document may have been altered from the original

Examination answers

1.1 Rings, polymers and analysis 1 (a) Structure:

Empirical formula: CH (b) Concentrated nitric acid and concentrated sulfuric

acid. (c) (i) NO2

+

(ii) HNO3 + H2SO4 → NO2+ + HSO4

– + H2O (iii)

�

NO2O2N

� H�

NO2H�

2 (a) (i) I: Br2, II: NaOH (ii) Dye, colouring or indicator. (iii) Add phenylamine to sodium nitrite and

hydrochloric acid below 10 °C. Add this product to an alkaline solution of phenol.

(b)

C C

C

CH

H H

H

H

C

C

H

C C

C

CH

H H

H

H

C

C

H

C C

CCC CCC

• Theporbitalsoverlapaboveandbelowtheplaneof the ring structure.

• Theelectronsaredelocalisedandspreadacrossall six carbon atoms.

• Inbenzeneallthebondsarethesamelengthbetween the lengths of the single and double bonds.

Phenolismorereactivebecausetheringisactivated.Thelonepairfromtheoxygenontheringis delocalised into the ring so electrophiles are more attractedtotheringthaninbenzene.

3 (a) (i) FeBr3 or AlBr3

(ii) Halogen carrier.

(iii)

� Br2 � HBr

Br

(iv) Bromobenzene (b)

Step 1 Step 2

Br

� H�

Br� H

Intermediate

Br

�

(c) (i)

H

HH

H H

HH

H H

H H

H H

H H

H

p orbitalsoverlap

Cyclohexene

Benzene

p orbitalsoverlap togive a� bondabove andbelow thetwo carbonatoms

See diagram00.00.00.00

H

HH

H H

HH

H H

H

(ii) Inbenzenetheπelectronsarespreadoverthewhole ring structure, as they are delocalised. When cyclohexene is reacted with bromine the bromine is polarised by the double bond. However,whenbenzeneisreactedwithbromine, the bromine cannot be polarised by theringastheelectronsarespreadoverthemolecule. Electrophiles are less attracted to the benzene.Greaterenergyisneededtobreaktheπcloudinbenzenethanincyclohexene.

4 (a) (i) D = Propanone (ii) E = Propanal (b) (i) Reagent(s):2,4-dinitrophenylhydrazine.

Observation:orangeprecipitate. (ii) Reagents:Tollens’reagent.ObservationforD:

nochange.ObservationforE:Silvermirrorformed.

2 © Pearson Education Ltd 2008This document may have been altered from the original

5 (a) CH3CHO + 2[H] → CH3CH2OH (b)

CH3 C

H

O

H

H

Organic product

O

CCH3

H

H�

��

��

NaBH4

Step 1

EthanalCH3 C

HH

O�

H O

H

��

Step 2

Intermediate

(iii) An electron-pair donor (iv) Thenucleophileisattractedtothepositive

charge.Lonepairofelectronsformsadativebond. An electron pair from the double bond goestotheoxygenatom.Theπ bond in the carbonyl bond breaks.

(c) Hydrogendoesnothavealonepairofelectrons. 6 (a)

O

Z-but-2-enalcis but-2-enal

E-but-2-enal(trans)

O

(b) (i) AddTollens’reagent.Heatreactioninawaterbath.But-2-enalgivesasilverprecipitateorsilvermirror.

(ii) Aldehydes can be oxidised but ketones cannot. (c) (i) CH3CH=CHCH2OH (ii) Redox reaction/reduction or addition. (d) C4H6O + 5O2 → 4CO2 + 3H2O 7 (a) (i) Reagent:Tollens’reagent.Observation:silver

mirror formed. Organic product: butanoic acid. (ii) Reagent:bromine.Observation:decolourises.

Typeofreaction:electrophilicaddition. (b) Recrystallise the product in order to purify the

orange precipitate. Measure the melting points of the crystals. Compare known melting points with data tables.

8 (a)

C

H

H

C

H

H

C OH

Butan-1-ol

H C

H

H H

H

C

H

H

C

CH3

H

C OH

2-methylpropan-1-ol

H

H

H

C

H

C

OH

C H

2-methylpropan-2-ol

H

H

CH HH H

H

C

H

H

C

H

O

C H

Butan-2-ol

H

H C

H

H H

H

(b) (i) Butan-1-ol can be oxidised to produce butanal andfurtheroxidisedtogivebutanoicacid.Thisoccurs as primary alcohols can be oxidised. Butan-2-ol can be oxidised to produce butanone. No further oxidation occurs as ketones cannot be oxidised. 2-Methylpropan-2-ol cannot be oxidised as it is a tertiary alcohol and is not oxidised. 2-Methylpropan-1-ol can be oxidised to produce 2-methylpropanal and further oxidised togive2-methylpropanoicacid.

D: 2-methylpropan-1-ol. E: 2-methylpropanoic acid. (ii) (CH3)2CHCOOH + C2H5OH →

(CH3)2CHCOOC2H5 + H2O (c)

CC

H

H

C

H

H

CHO C

H

H H

H

O

O�Na�

9 (a) (i) Methyl butanoate. (ii) Heat/boil/warm/reflux with aqueous HCl or

aqueous NaOH. (b) (i)

HO

(ii)

or

C2H5

HOCH2CH2 H

H

HO

C C


(iii)

O

CH2CH3CH2

C C

H

CH2

H

C

O

H3C

10 (a) (i) Concentrated H2SO4. (ii) Thisistopreventthelossofreactantsand

productsbyevaporation. (b) (i)

C

O

C HH

C

H

HH

C

H

HH

C O C HH C

H

H

(ii) Butan-2-ol. (c) Flavouringsorperfumes.11 (a) A compound containing nitrogen where the nitrogen

is attached to one carbon atom only. (b) C2H5NH2 + H+ → C2H5N

+H3

(c) (i) StageI:Sn,conc.HCl.StageII:NaNO2, HCl. StageIII:Phenol,NaOH.

(ii)

N N

NaOH

Stage I

Stage III

CH3 OH � NaCl � H2O

� 6[H]

NO2

CH3

� 2 H2O

NH2

CH3

�

OHN2�Cl�

CH3

(iii) Dyes.

12 (a) Stage 1: React phenylamine with sodium nitrite and hydrochloricacidat10ºCorbelow.Thisformsadiazoniumsaltwhichisunstableandmustbekeptbelow10ºC.Stage2:Reactthediazoniumsaltwithan alkaline solution of phenol below 10 ºC.

(Diazonium salt)

� 2 HCI � NaNO2 � NaCl � 2 H2O

N2�Cl�NH2

NaCl � H2O

NaOH

HO

�

N2�Cl� OH

N N

(b) (i)

E110SO3

�Na�

N

N SO3�Na�

OH

(ii) 16 carbon atoms and 10 hydrogen atoms. (c) NaOH(aq) (d)

H2N

Phenol

Amine

SO3�Na�

SO3�Na�

OH


13 (a) Diamino – compound contains two amine groups 1,4 – shows the position of the amine groups on the benzenering.

(b) (i) Reduction/redox reaction. (ii) TinandconcentratedHCl.Heatunderreflux. (iii)

NO2 � 12[H]

H2N

O2N

NH2 � 4H2O

(c) (i) TheaminogroupacceptsH+ using its lone pair or electrons on the nitrogen and forming a co-ordinate bond.

(ii)

Cl�H3N� �NH3Cl�

14 (a)

CH2

C C

H

CH3(CH2)6CH2

H

(CH2)6COOH

(b) (i) Oleic acid: C18H34O2. Ethanol: C2H6O. Ethyl oleate: C20H38O2

(ii) C18H34O2 + C2H6O → C20H38O2 + H2O (c) Hydrolysis using hot aqueous HCl.

1.2 Polymers and synthesis 1 (a) (i) Hydrochloric acid. Aqueous acid heated under

reflux. (ii) Amino acids. (iii)

CC

H

H

H N

H

H

(or equivalent)

O

O H

�

(iv) A reaction with water which breaks down a compoundintosmallersub-units.Inthereactionabovethepeptidebondisbroken.

(b) (i) For stereoisomerism in compound A, there must be a chiral centre, which is a carbon atom with four different groups attached to it. Stereoisomerism occurs when there is a differentspatialarrangementofthegroups.Twomirror images that are non-superimposable are observed.

(ii) Themoleculecontainstwochiralcentresandeach chiral centre has two isomers.

(iii) Use a naturally occurring amino acid. (iv) Ahigherdosewouldberequired.Theother

stereoisomermayhavesideeffects. 2 (a) (i) RCH(NH2)COOH (ii)

C

O

O

C

CH H

C

H

C C H

H

H

HH H

H

N

H H

H

(b) (i) Anionwhichhasbothapositiveandanegativecharge.Theoverallchargeiszero.

(ii) A proton is transferred from the acidic COOH group to the basic NH2 group.

C

O

O

C

R

H

N

H H

H

C

O

O�

C

R

N�

H

H H

H

(c) (i) Heat with HCl (aq). (ii) Hydrolysis. (d) (i) Ammonia in ethanol. (ii) Leucine synthesised in the laboratory contains a

mixture of two optical isomers. Leucine from meat (a natural source) contains only one of the isomers.

3 (a) For L:

H

H

H

C C

For M:

C

O HO

CO

H C

OH

H

C

H

OH

H

H O

(b)

O C

O

O C

O

O C

O


(c) Itisacondensationreaction.Whenthereactionoccurs a small molecule such as water is lost.

(d) Fibres/clothing. 4 (a) (i)

H HHHHH

H3C

CC

CCC C

CH3 H3C CH3 H3C CH3

(ii) CH2

(iii)

H

C C

CH3

H

H

(b) (i) Name of functional group: peptide.

C

O

N

H

(ii) Condensation. (iii) Displayed formula of H:

O

C CC

H

H

NC

H

H

N

H O

O HH

H

Skeletal formula of H:

N

H

OH

O

H2N

O

(iv) No. of moles of glycine = 1.40/75.0 = 0.0187 moles. Expected no. of moles of dipeptide = 0.0187/2 = 9.33 × 10–3 mol. Expected mass = 9.33 × 10–3 × 132 = 1.232 g. % = (1.10/1.232) × 100. Percentage yield 89.3%

(v)

H CC

HH

H

NCl�

H O

OH

�

5 (a) Inadditionpolymerisation,theC=Cdoublebondinan unsaturated molecule (monomer), such as ethene, breaks open and the monomer adds on to agrowingpolymerchain.Themonomersaddoneat a time to form a long polymer such as poly(ethene), HDPE.

Incondensationpolymerisation,apolymersuchasPETisformedwhenmonomers–inthiscaseethane-1,2-diolandbenzene-1,4-dicarboxylicacid– join together and lose a small molecule for each bondthatforms.Inthecaseofthispolyester,awater molecule forms for each ester linkage that forms between –OH and HOOC–.

C

O

HO

H

H

H

C C

H

H

Cn C

H

H

Addition: Monomer ethene

H n

H

H

H

C C

H

H

H

O C

n

H

C O

H

Condensation

C

H

H

C OH �

H

H

C

O

C

O

C

O

OHHO

� H2O

(b) (i)

O

C C

H

H

H

C O

CH3

(ii) Reagents: HCl (aq). Conditions: Heat and reflux (iii) CH3COOH (c)

O

C

�

HO C

H

H

C

H

H

OH

OH

O

CHO

6 (a)

HCIN

H

O

C

Cl

(CH2)4 C

O

Cl

H

N

H

(CH2)6 N

H

HHexanedioyl dichloride

Repeat of unit polymer

�

�

1,6-diaminohexane

Otherproduct

C

O

(CH2)6(CH2)4 NC

O

H


(b) (i)

�

O

COH

O

CHO

H

NH

H

NH

(ii) TheintermolecularforcesinNomex®arestronger than in nylon.

7 (a)

NO2

(b) CH3COOH and CH3OH (c)

C

O

OH

O

C

O

CC OH

C6H5

H

NC

H

CH3

N

HH

H

O

C C

C6H5

H

NC

H

CH3

N

HH

H

8 (a) (i) NaOH (aq) (ii) Substitution. (iii) C6H5CH2Cl + NaOH → C6H5CH2OH + NaCl (b) (i)

O

C C

H

H

OC

H

H

H

(ii) C6H5CH2OH + CH3COOH → CH3COOCH2C6H5 + H2O

(iii) Flavourings.

(iv)

C

O

ClC��

H

H

C

H

H

HO

C

O

O

H

H

CCH3 � Cl�

�

9 (a) C10H10O5

(b) (i) Reagent(s): K2Cr2O7 (aq). Conditions: heat with H+. Organic product:

CH2CH2 C C

OO

OHOC

O

H

(ii) Reagent(s): HCl (aq). Conditions: aqueous. Organic products:

CH2CH2 C C

OO

OHHOHO OH

(iii) Reagent(s): NaOH (aq). Conditions: none. Organic product:

CH2CH2 C C

OO

O�Na�OHO

10 (a) HOOC(CH2)4COOH and H2N(CH2)6NH2. (b) condensation polymerisation – small molecule, H2O,

is eliminated. (c) Structural similarity Bothhavepeptideoramidelinks Both form H-bonds between molecules. Chemical similarity Both can be hydrolysed back to monomers by

reaction with hot aqueous acid or alkali: e.g. –CO(CH2)4CONH(CH2)6NH– →

HOOC(CH2)4COOH + H2N(CH2)6NH2 Differences Protein can be water-soluble, nylon-6,6 not. Proteins are biodegradable, nylon-6,6 is not. Nylon-6,6 has a regular chain of alternating

monomers. Proteins are made from 20 different amino acids.


11 (a) (i)

CH3

H

H

C C

H

CH3

C C

H

H

H n

(ii) Addition:ThemonomerhasaC=Cdoublebond.Thedoublebondbreaksandnosubstance is lost.

Condensation: A small molecule such as water is lost.

(b)

CH2

CH3

Cl

O CH2

CH3

O

O O

CH CHC

(c) (i)

CH2

CH3

Cl

O CH2

CH3

O

O O

CH CHC

(ii) H reacts with NaOH, poly(propene) does not. H is an ester and is polar so will be hydrolysed

by NaOH. Poly(propene) is non-polar. 12 (a)

O2N N

N OH

COOH

OH

COOH

NaOH

O2N N

N

COO�Na�

OH�Na�

O2N N

N OH

COOH

O2N NH2

O2N N2�Cl�

O2N N

N

OH � 2 H2O

COOH

H2N N

N

OH � 6[H]

COOH

Below 10 °C

NaNO2/HCI

(b)

O2N N

N OH

COOH

OH

COOH

NaOH

O2N N

N

COO�Na�

OH�Na�

O2N N

N OH

COOH

O2N NH2

O2N N2�Cl�

O2N N

N

OH � 2 H2O

COOH

H2N N

N

OH � 6[H]

COOH

Below 10 °C

NaNO2/HCI

(c)

O2N N

N OH

COOH

OH

COOH

NaOH

O2N N

N

COO�Na�

OH�Na�

O2N N

N OH

COOH

O2N NH2

O2N N2�Cl�

O2N N

N

OH � 2 H2O

COOH

H2N N

N

OH � 6[H]

COOH

Below 10 °C

NaNO2/HCI

(d) Tinandconcentratedhydrochloricacid,reflux.

O2N N

N OH

COOH

OH

COOH

NaOH

O2N N

N

COO�Na�

OH�Na�

O2N N

N OH

COOH

O2N NH2

O2N N2�Cl�

O2N N

N

OH � 2 H2O

COOH

H2N N

N

OH � 6[H]

COOH

Below 10 °C

NaNO2/HCI

1.3 Analysis1 (a) (i) Paper, column or thin-layer chromatography. (ii) TheRfvalue. (iii)

x

yRf � x

y

Solvent front

Start point

(b) (i) Retention time. (ii)

Retention time

Dete

ctio

n

Start

(c) (i) Relativesolubilityinliquidstationaryphaseoradsorption to solid stationary phase.

(ii) Each component distributes itself between the mobile phase (carrier gas) and the stationary phase.

Different components are held more or less strongly on the stationary phase and this leads to separation.

Thisdependsuponmolecularsize,volatilityandextent of bonding to the stationary phase.


2 (a) Adsorption at 1700 cm–1 shows the presence of C=O. Adsorption at 1200 cm–1 shows the presence of C–O. No broad adsorption at 2500–3500 cm–1 so does not

contain O–H. (b) ThemassspectrumshowsanM+ peak at 116 therefore 116 – 32 (2 oxygen atoms) = 84 therefore 6 carbon atoms maximum. Molecular formula of X = C6H12O2. Base peak = 57 CH3CH2C=O. NMR. peak areas show 12H atoms and four different

protonenvironments. δ = 0.9 ppm suggests CH3 next to CH2 (triplet) δ = 1.2 ppm suggests 2 × CH2 next to CH (as split

into a doublet). δ = 2.3 ppm CH2 next to CH3 (quartet) Other CH2 next to C=O δ = 4.1 ppm is CH next to 2 × CH3.

C

C

H

CH3

CH3

C

O

O

C

H

H

H

H

H

3 (a) Compound has a molecular mass of 74. Relativeempiricalformulaemass=74

(3 × 12 + 6 + 2 × 32) therefore empirical formula = molecular formula

= C3H6O2.

100

80

60

40

20

010 15 20 25 30 35 40 45 50 55 60 65 70 75

m/e

Rela

tive

inte

nsity

(b) (i) Compound X is not an aldehyde or a ketone. (ii) Compound X does not contain C=C and is not a

phenol. (c) Structure 1: ethyl methanoate. Structure 3: propanoic acid. (d) Peak at 1750 cm–1 indicates presence of C=O. Peak at 1250 cm–1 indicates presence of C–O. No peak at 2500–3500 cm–1 means it does not

contain O–H group. Itcannotbestructure3.

(e) Correct structure of compound Y:

O

C HC

H

H

OC

H

H

H

Peak at δ = 2.1 ppm is due to –CO–CH3

Peak at δ = 3.7 ppm is due to –O–CH3

Relativepeakareas1:1asbothgroupshavethesame number of carbon atoms.

Peaks show no splitting as there are no protons on the neighbouring carbon atoms.

4 (a) Low boiling point. (b) Add2,4-DNP(2,4-dinitrophenylhydrazine)tothe

aldehyde or ketone. Purify or recrystallise the orange precipitate. Measure the melting point of the crystals. Comparetheresultswithknownvaluesfromadatatable.

(c) (i)

H

HC

C

C

C

H

H

H

HH

OH

O

Abso

rptio

n of

ene

rgy

11 10 9 8 7 6 5 4 3 2 1 0Chemical shift/�

n.m.r. spectrum of 3-hydroxybutanone

1 1 3 3

(ii) Re-runthespectrumhavingaddedD2O to the sample.Thepeakdueto–O–Hwilldisappear.

(iii) Peak at δ = 1.4 ppm (doublet): (1:1) due to one proton on adjacent carbon atom. peak at δ = 4.3 ppm (quartet): (1:3:3:1) due to three protons on the adjacent

carbon atom. (iv) See graph answer to (c)(i). (v) Thenumberofprotonsinthesamechemical

environment.5 (a) (i)

C

O

C

O

HO

C

H

H

C*

OH

H OH

(ii)

CH2COOH

CCOOHH

HO

CH2COOH

CHHOOC

OH


(b) (i)

100

80

60

40

20

040 60 80 100 120 140

Rela

tive

abun

danc

e (%

)

m/e

(ii) Molecular ion peak would be at 90. (c) (i)

Abso

rptio

n of

ene

rgy

12 11 10 9 8 7 6 5 4 3 2 1 0Chemical shift/�

1

3

(ii) 4asOHprotonswouldnowbevisible. (d) (i)

C

H

H

H

C

C

H

H

C

O

O

C

O H

H

H

H

H

(ii) More fruity because of ester/less sour as acid is used up.

6 (a)

C

H

C

H

C HO

H

CH HH H

H

O

CC C

H

H

H

H

H

(b) CH3CH2COOH + (CH3)2CHCH2OH → CH3CH2COOCH2CH(CH3)2 + H2O

Alcohol: 2-methylpropan-1-ol Carboxylic acid : propanoic acid. Heat the alcohol and the carboxylic acid with some

concentrated sulfuric acid in a water bath. (c) Mass spectrometry. Use the molecular ion peak, the m/zvalueforthe

peak at highest mass. (d) (i) Theδvalueorchemicalshiftindicatesthetypeof

protonorchemicalenvironment.Thenumberofpeaksgivesthenumberofdifferenttypesofproton or the different numbers of chemical environments.Therelativeratioofpeakareasshows the number of protons of each type. Splittingpatternsgiveinformationaboutthenumber of neighbouring hydrogen atoms

(protons)onthecarbonunderinvestigation:thesplitting pattern (singlet, doublet, triplet or quartet) is one greater than the number of adjacent protons.

D2O is used to identify the presence of O–H groups.

(ii)

CH3CH2COOCH2CH(CH3)2

S Q P R T

7 (a) (i) Find the m/zvalueofthepeakfurthesttotherightwith highest m/zvalue.

(ii) C2H3O2 has an empirical mass of 59. therefore 118/59 = 2 2 × empirical formula = C4H6O4. (b) (i) OH peak disappears. (ii) Peak at δ=3.3ppmidentifiesCHgroup.It

shows as a quartet splitting pattern showing presence of three protons on the adjacent carbon atom.Thereisonlyonehydrogeninthisenvironment.

Peak at δ = 1.2 ppm identifies CH3group.Itshows doublet splitting showing the presence of oneprotonontheadjacentcarbonatom.Therearethreeprotonsinthisenvironment.

Relativepeakareasshowthenumberofhydrogens(protons)intheenvironment.

C

H

H

CH H

C

O

C

O

HO OH

(iii) Therewouldbeonepeakwithnosplittingasallfourprotonsareinthesameenvironment.

8 (a) Infrared: X and YbothhaveC=Oat1680–1750cm–1. X and YbothhaveC–Oat1000–1300cm–1. X and YbothhaveO–Hinrange2500–3300cm–1. Only Y has an absorption at 3100–3500 cm–1 for N–H. Bothhavedifferentfingerprintregions. Mass spectrum: X and YhavedifferentM peaks, X = 88 and Y = 89. X and Yhavesomesimilarfragments,e.g.CH3CH+

at m/z = 28, COOH+ at 45, CHCOOH+ at 58, CH3CHCOOH+ at 73.

X and Yhavesomedifferentfragments,e.g.X has CH3CHCH3

+ at m/z = 43 or (CH3)2CHCO+ at 71; Y has H2NNCH+ at 29, H2NCHCH3

+ at 44, CH3(NH2)CHCO+ at 72 or NH2

+ at 16.


(b) Proton environment

Chemical shift, δ/ppm

Splitting pattern

Relative peak area

–O–CH3 3.3–4.3 singlet 3

–CH2–CO 2.0–2.9 quartet 2

CH3–C– 0.7–1.6 triplet 3

2.1 Rates, equilibrium and pH1 (a) H2O2+2I– + 2H+ →I2 + 2H2O (b) (i) Experiment2hastwice[I–] of Experiment 1 and

the rate has quadrupled, soorder=2withrespecttoI–. Experiment 3 has twice [H+] of Experiment 2 and

the rate is unchanged, so order = 0 with respect to H+. (ii) rate = k [H2O2][I–]2

(iii) k = rate/[H2O2][I–]2

Experiment 1: k = 1.15 × 10–6/(0.01)(0.01)2 = 1.15 × 10–2 dm6 mol–2 s–1

(c)

Rate

of r

eact

ion

[H2O2(aq)]/mol dm�30

Straight line increasingThrough 0/0

(d) (i) 2H2O2 → 2H2O + O2

(ii) 1 dm3 H2O2 → 20 dm3 O2

amount of O2 = 20/24 mol concentration of H2O2 = 2 × 20/24

= 1.67 mol dm–3

2 Fromthegraph,thereisaconstanthalf-life.Therefore,order = 1 with respect to [CH3COCH3].

From the table, the rate doubles when [H+] doubles. Thereforeorder=1withrespectto[H+].

Fromthetable,theratestaysthesamewhen[I2] doubles. Thereforeorder=0withrespectto[I2].

Therateequationis:rate = k[H+][CH3COCH3]

k = rate _______________ [H+][CH3COCH3]

= 2.1 × 10–9

_______________ 0.02 × 1.5 × 10–3

= 7.0 × 10–5 dm3 mol–1 s–1

Therate-determiningstepinvolvesthespeciesintherateequation: H+ and CH3COCH3.

A possible two-step mechanism is: Step 1: CH3COCH3 + H+ → CH3COHCH3

+

Step 2: CH3COHCH3++I2 → CH3COCH2I+HI+H+

H+ is used up in the first step and is regenerated in the second step so H+ is a catalyst.

3 (a) Kc = [CH3COOC2H5] [H2O]

___________________ [CH3COOH] [C2H5OH]

(b) (i) CH3COOH, 1.0 mol; C2H5OH, 7.5 mol CH3COOC2H5, 5.0 mol; H2O, 5.0 mol

(ii) Kc = 5 × 5 _______ 1 × 7.5

= 3.3 (no units)

(c) Theexperimentcouldbeleftlongertoreachequilibrium.

Thecompositionsofthemixturescouldbeanalysedattimeintervalsuntilthecompositionsdonotchangeany more.

(d) (i) TherewouldbemoreCH3COOC2H5 and H2O TherewouldbelessCH3COOH

Theequilibriumshiftstotherighttocounteractthe effect of increasing the ethanol concentration.

(ii) Kc stays the same. (e) Thecompositionstaysthesamebecausethecatalyst

does not shift equilibrium position. Theratesoftheforwardandreversereactionsare

increased by the same amount. (f) (i) A decrease in Kc results in more reactants and

fewerproducts.TheequilibriumthereforeshiftstotheleftinresponsetothenewvalueofKc.

(ii) Increasingthetemperaturemovestheequilibriumto the left to counteract the effect of the increase inenergy.Theforwardreactionisexothermicandthereversereactionisendothermic.

4 (a) Kc = [Hl]2

______ [H2] [l2]

(b) (i) H2,0.14mol;I2,0.04mol;HI,0.32mol

(ii) Kc = 0.322

___________ 0.14 × 0.04

= 18.285 714 29

= 18 (to 2 sig. figs), no units (c) Kcisconstantbecauseallvolumescancelandthe

temperature is constant. Thecompositionisthesameastherearethesame

number of gas moles on either side of the equation. (d) Theforwardreactionisexothermic.Theequilibrium

movestothelefttocompensatefortheincreaseintemperature.

(e) (i) I2(aq) + H2S(g) →2HI(aq)+S(s) (ii) amountofI2 reacted = 1.89 mol, ThereforeamountofHIformed=3.44mol TheoreticalamountofHIproduced=3.78mol

% yield = 3.44 × 100 __________ 3.78

= 91.0%


(iii) [HI]=3.44 × 1000 ___________ 750

= 4.59 mol dm–3

pH = –log 4.59 = –0.665 (a) (i) A Brønsted–Lowry acid is a proton donor. (ii) A weak acid is partially dissociated. (iii) pH = –log[H+] (iv) A buffer solution minimises changes in pH after

addition of small amounts of both acids and alkalis.

(b) IntheequilibriumH2CO3(aq) Y H+(aq) + HCO3–(aq)

Hydrogencarbonate, HCO3–, reacts with any added

acid: HCO3

– + H+ → H2CO3

Theequilibriummovestothelefttocounteractincreased [H+].

H+ reacts with any added alkali: H+ + OH– → H2O Theequilibriummovestotherighttocounteracta

decrease in H+.

(c) Ka = [H+][HCO3

–(aq)] _____________

[H2CO3(aq)]

[H+] = 10–pH = 10–7.4 = 3.98 × 10–8 mol dm–3

[HCO3

–(aq)] __________

[H2CO3(aq)] =

Ka ____ [H+]

= 4.17 × 10–7

__________ 3.98 × 10–8

= 10.5

6 (a) A weak acid partially dissociates: e.g. HCOOH Y H+ + HCOO–

(b) (i) pH = –log (1.55 × 10–3) = 2.81 [H+]dealswithnegativeindicesoveraverywide

range.

(ii) Ka = [H+(aq)][HCOO–(aq)]

_________________ [HCOOH(aq)]

(iv) Ka = [H+(aq)]2 ____________

[HCOOH(aq)] =

(1.55 × 10–3)2 ____________

0.015

= 1.60 × 10–4 mol dm–3

pKa = –log Ka = –log (1.60 × 10–4) = 3.80

(v) Percentage dissociating = (1.55 × 10–3) × 100

_________________ 0.015

= 10.3% (c) (i) HCOOH + NaOH → HCOONa + H2O (ii) n(HCOOH) = 0.0150 × 25.00/1000

= 3.75 × 10–4 mol From the graph, the rapid rise in pH takes place

after addition of 30 cm3ofNaOH(aq).Therefore30cm3 of NaOH(aq) are required for neutralisation.

Therefore[NaOH]=3.75×10–4 × 1000/30 = 0.0125 mol dm–3

(iii) Kw = [H+(aq)][OH–(aq)] pH = –log(1 × 10–14/0.0125) = 12.10

(iv) metacresol purple Theindicator’spHrangecoincideswiththepH

change during the sharp increase in pH in the titrationcurve.

7 (a) Theextentofdissociationoftheacid (b) (i) H2SO3(aq) + CH3COOH(aq) Y

HSO3–(aq) + CH3COOH2

+(aq) acid 1 base 2 base 1 acid 2 (ii) CH3COOH is a stronger acid than C6H5OH.

Therefore,CH3COOH will donate H+ to C6H5OH CH3COOH(aq) + C6H5OH(aq) Y

CH3COO–(aq) + C6H5OH2+(aq)

(c) For HCl, pH = –log[H+] = –log 0.045 = 1.35 For CH3COOH, [H+] = √

________________ (Ka × [CH3COOH])

[H+] = 8.75 × 10–4 mol dm–3

pH = –log 8.75 × 10–4 = 3.058 (d) 2 mol of each acid produce 1 mol of H2

Mg + 2HCl → MgCl2 + H2

Mg + 2CH3COOH → (CH3COO)2Mg + H2

HCl is the stronger acid and in solution [H+] in HCl > [H+] in CH3COOH.

HCl will therefore react quicker as its [H+] is greater.8 (a) (i) Kw is the ionic product of water (ii) Kw = [H+(aq)] [OH–(aq)]

(b) amount of HCl = 5.00 × 10–3 × 21.35 __________________ 1000

= 1.067 × 10 –4 mol

amount of Ca(OH)2 = 1.067 × 10–4

___________ 2

= 5.34 × 10–5 mol concentration of Ca(OH)2 = 40 × 5.34 × 10–5

= 2.136 × 10–3 mol dm–3

(c) [OH–] = 2 × 2.7 × 10–3 = 5.4 × 10–3 mol dm–3

[H+(aq)] = Kw ________

[OH–(aq)] = 1.0 × 10–14

__________ 5.4 × 10–3

= 1.85 × 10–12 mol dm–3

pH = –log (1.85 × 10–12) = 11.73/11.7 (d) [H+] [OH–] = 10–14 mol2 dm–6; If[OH–] is 100 (102) × greater than [H+], [OH–] must be

10–6 and [H+] must be 10–8

pH = 10–pH, so pH = 89 (a) (i) a weak Brønsted–Lowry acid is a proton donor

that partially dissociates

(b) Ka = [HCOO–][H+]

___________ [HCOOH]

= [H+]2 _________

[HCOOH]

[H+] = √_____________________

{(1.58 × 10–4) × (0.025)} = 1.99 × 10–3 mol dm–3

pH = –log[H+] = –log 1.99 × 10–3 = 2.70


(c) (i) A buffer solution minimises pH changes. (ii) sodium methanoate, HCOONa HCOONa supplies HCOO– as the conjugate

base. (iii) ThepHofthebuffersolutiondependsuponthe

extentofdissociationgivenbytheKavalueandtherelativeconcentrationsoftheweakacidandconjugate base.

(d) Mass of HNO3 = 1400 × 65 __________ 100

/ 910 g

amount of HNO3 = 910 ____ 63

= 14.4 mol

pH = –log[H+] = –log 14.4 = –1.16

(e) 2HNO3 + CaCO3 → Ca(NO3)2 + CO2 + H2O (f) A conjugate acid–base pair is two species differing

by H+

acid 1: HNO3; base 2: NO3–

acid 2: HCOOH2+; base 2: HCOOH

(g) (i) 6HNO3 + S → H2SO4 + 6NO2 + 2H2O or 4HNO3 + S → H2SO4 + 4NO2 + H2

(ii) TheoxidationnumberofNdecreasesfrom+5inHNO3 to +4 in NO2.

2.2 Energy 1 (a) (i) Energyisneededtoovercometheforceofattractionbetweenouterelectronsandnucleus (ii) Electronaffinityinvolvesaneutralatomgainingofanelectronbyattractionfromthenucleus (b) (i)

0

�Hf

CaO(s)

Ca(s) � ��O2(g)

Ca(g) � ��O2(g)

Ca(g) � O(g)

Ca�(g) � O(g) � e�

Ca2�(g) � O(g) � 2e�

�HEA1 � �798 kJ mol�1

�HLE � �3459 kJ mol�1

�HEA1 � �141 kJ mol�1

�HI2 � �1150 kJ mol�1

�HI1 � �590 kJ mol�1

�Ha � �249 kJ mol�1

�Ha � �178 kJ mol�1

Ca2�(g) � O�(g) � e�

Ca2�(g) � O2�(g)

(ii) ∆Hf = +178 + 249 + 590 + 1150 + (–141) + 798 + (–3459) = –635 kJ mol–1

(iii) LatticeenthalpyofFeOismorenegativeandmore exothermic than lattice enthalpy of CaO. IonicradiusofFe2+ must be less than that of a Ca2+ion.Thisleadstogreaterattraction between Fe2+ ions and O2– ions than between Ca2+ ions and O2– ions.


2 (a) Lattice enthalpy is the enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions under standard conditions.

Mg2+(g) + 2Cl–(g) → MgCl2(s) (b)

Mg2�(g) � 2Cl(g) � 2e�

Mg�(g) � 2Cl(g) � e�

Mg(g) � 2Cl(g)

Mg(g) � Cl2(g)

Mg(s) � Cl2(g)

MgCl2(s)

Mg2�(g) � 2Cl�(g)

�HLE

2 � �349�HI2 � �1451 kJ mol�1

�HI1 � �738 kJ mol�1

2 � �Ha � 2 � �123 kJ mol�1

�Ha � �148 kJ mol�1

�Hf � �641 kJ mol�1

0

–641 = +148 + 2 × 123 + 738 + 1451 + 2 × (–349) + ∆HLE

∴ ∆HLE = –641 – (+148 + 2 × 123 + 738 + 1451 + 2 × (–349)) = –2526 kJ mol–1

(c) Na+ has a larger radius than Mg2+ and Br– has a larger radius than Cl–

Na+ has a smaller charge than Mg2+

Br– experiences less attraction than Cl–

Na+ experiences less attraction than Mg2+

ThereforethereismuchlessattractionbetweenNa+ ions and Br– ions than between Mg2+ ions and Cl– ions. One mark for correct spelling, punctuation and grammar in at least two sentences.

3 (a) (i) ∆H1 = Enthalpy change of formation of magnesium oxide

∆H2 = Enthalpy change of atomisation of magnesium

∆H3 = First ionisation energy of magnesium (ii) Mg2+(g) + O2−(g) (iii) Theelectronbeinggainedisrepelledbythe

negativechargeoftheO–(g) ion. (b) (i) –602 = 149 + 736 + 1450 + 248 + 650 + ∆HLE

∴ ∆HLE = –602 – (149 + 736 + 1450 + 248 + 650) = −3835 kJ mol−1

(ii) Lattice enthalpy of barium oxide is less exothermicthanthatofmagnesiumoxide.Thisis because Mg2+ has a smaller ionic radius than Ba2+ so there is a stronger attraction between thepositiveandnegativeions.

(c) Magnesium oxide has a high melting point because it has a highly exothermic lattice enthalpy resulting fromstrongattractiveforcesbetweenMg2+ and O2– ions.

4 (a)

NH4�(g) � Cl�(g)

NH4�(aq) � Cl�(g)

NH4�(aq) � Cl�(aq)

NH4Cl(s)

Hydration of Cl–

Hydration of NH4�

Latticeenthalpy

Enthalpychange ofsolution

(b) ∆Hhyd(NH4+) + ∆Hhyd(Cl–) = ∆HLE(NH4Cl) + ∆Hs(NH4Cl)

∆Hhyd(NH4+) + (–364) = –689 + 15

Rearranging: ∆Hhyd(NH4+) = (+364) + –689 + 15

∴ ∆Hhyd(NH4+) = –310 kJ mol–1

(c) Thereisanincreaseindisorderondissolving. Increaseinentropyoutweighsincreaseinenthalpy. 5 (a) Energy lost = mc∆T = 108.48 × 4.18 × 12.5

= 5668 J = 5.668 kJ

AmountofLiCldissolved=m __ m = 8.48 ___________ (6.9 + 35.5)

= 0.200 mol

∴ ∆Hsolution = – 1.60 ______ 0.200

= –28.34 kJ mol–1


(b) Thereisanincreaseindisorderondissolving. Increaseinentropyoutweighsincreaseinenthalpy. (c) (i) X: Lattice enthalpy of LiCl; Y: Enthalpy change

of hydration of Li+

(ii) Li+ ions are smaller than Na+ ions. Small ions exert more attraction on water

molecules and more energy is released. 6 (a) InEquation1,∆Sis+vebecauseagasisreleased. InEquation2,∆Sis–vebecausethenumberofgas

molecules decreases. (b) ∆G = ∆H – T∆S. (c) (i) Increaseinentropyfromformationofagas

outweighs increase in enthalpy. Overall,balancebetweenenthalpyandentropy

is such that ∆H – T∆S < 0. (ii) At high temperatures, decrease in entropy from

decrease in the number of gas molecules outweighs decrease in enthalpy because T∆S becomes more significant.

Overall,balancebetweenenthalpyandentropyis such that ∆H – T∆S > 0.

7 (a) ∆Sispositiveastheforwardreactionresultsinanincrease in the moles of gas molecules.

(b) ∆S = ΣSØ (products) – ΣSØ (reactants) = (198 + 3 × 131) – (186 + 189) = +216 J K–1 mol–1

(c) ∆G = ∆H – T∆S At room temperature, ∆G = +210 – 298 × 0.216

= +145.632 kJ mol–1

For a reaction to take place, ∆G<0kJmol–1

(d) When ∆G =0, 0 = ∆H – T∆S

∴ T = ∆H ___ ∆S

= 210 ______ 0.216

= 972 K = 699 ºC

8 (a) Thestandardelectrodepotentialofahalfcell,EØ, is the e.m.f. of a half cell compared with a standard hydrogen half cell, measured at 298 K with solution concentrations of 1 mol dm–3 and a gas pressure of 100 kPa (1 atmosphere).

(b) (i)

Cu(s)

Ag�(aq)Cu2�(aq)

Ag(s)

V

Salt bridge

Electrons

(ii) Direction from Cu(s) to Ag(s) shown near to wire (iii) Standard cell potential = 0.80 – 0.34 = 0.46 V

(iv) Cu + 2Ag+ → Cu2+ + 2Ag (c) EØ for Fe3+/Fe2+islesspositivethanforCl2/Cl–. ThereforetheFe2+ has a greater tendency to lose

electrons than Cl–. EØforI2/I–islesspositivethanforFe3+/Fe2+. ThereforetheI– has a greater tendency to lose

electrons than Fe2+. 9 (a) Thestandardelectrodepotentialofahalfcell,EØ,

is the e.m.f. of a half cell compared with a standard hydrogen half cell, measured at 298 K with solution concentrations of 1 mol dm–3 and a gas pressure of 100 kPa (1 atmosphere).

(b) Standard cell potential = 1.52 – 1.36 = 0.16 V (c) (i) 2MnO4

– + 10Cl– + 16H+ → 2Mn2+ + 5Cl2 + 8H2O

(ii) Cl from –1 to 0 Mn from +7 to +2 (iii) Cl– has been oxidised Mn in MnO4

– has been reduced (d) Standard electrode potential applies only to a

concentration of 1 mol dm–3,andpotentialsgivenoinformation about rates. When concentration of Cl–decreases,equilibriummovestowardsCl–, and Cl2/Cl–halfcellpotentialbecomeslessnegative.Thedifference between the electrode potentials becomes smaller and the Cl2/Cl–potentialmayevenbecomemorepositivethantheMnO4/Mn2+ potential.Thereactionwillthennotbefeasible.

10 (a) (i)

Cr(s)

Cd2�(aq) (1 mol dm�3)Cr3�(aq) (1 mol dm�3)

Cd(s)

T � 298 KV

Salt bridge

Electrons

(ii) Direction from Cr(s) to Cd(s) shown near to wire (iii) Cr half cell EØvalueismorenegative. (b) 2Cr(s) + 3Cd2+(aq) → 2Cr3+ (aq) + 3Cd(s) (c) (i) Standard cell potential = –0.40 – (–0.74) = 0.34 V (ii) Concentration of Cr3+ decreases so equilibrium

movestowardsCr3+ and Cr half-cell becomes morenegative.Thereforethereisabiggerdifference between the electrode potentials and the cell potential will increase.


11 (a) Thestandardelectrodepotentialofahalf-cell,EØ, is the e.m.f. of a half-cell compared with a standard hydrogen half-cell, measured at 298 K with solution concentrations of 1 mol dm–3 and a gas pressure of 100 kPa (1 atmosphere).

(b) (i) solution A = 1 mol dm–3 HCl(aq) (or 1 mol dm–3 H+(aq))

solid B = platinum electrode (ii) Direction from hydrogen half-cell along wire to

S2O82– half-cell

(iii) S2O82– + H2 → 2SO4

2– + 2H+

(c) molar mass of Na2S2O8 = 238.2 g mol–1

molar mass of Na2SO4 = 142.1 g mol–1

∴ 23.82 g of Na2S2O8 and 14.21g of Na2SO4

(d) Concentration of SO42– decreases so equilibrium

movestowardsSO42– and S2O8

2–/SO42– half-cell

becomesmorepositive.Thereforetherewillbeabigger difference between electrode potentials and the cell potential will increase.

12 (a) (i) moles of Cu = 0.68 × 5/1000 = 0.0034 mass of Cu = 0.0034 × 63.5 = 0.216 g % Cu = 0.216/0.28 = 77% (ii) Ratios Cu : N : O : H = 26.29/63.5 : 11.6/14 : 59.63/16 : 2.48/1 = 0.414 : 0.829 : 3.73 : 2.48 = 1 : 2 : 9 : 6 Empirical formula of A = CuN2O9H6 (iii) Formula usually shown as: Cu(NO3)2·3H2O (b) Cu → Cu2+: Cu from 0 to +2 NO3

– → NO: N from +5 to +2 ratio of Cu : N in equation is 3 : 2 3Cu + 8H+ + 2NO3

– → 3Cu2+ + 2NO + 4H2O (c) (i) moles of A = 90/24000 = 3.75 × 10–3

Mr of A = 0.24/ 3.75 × 10–3 = 64 GasisSO2

(ii) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O

2.3 Transition elements 1 (a) Addaqueoussodiumhydroxidewhichgivesarusty

brown precipitate. (b) (i) Cr2O7

2– + 14H+ + 6Fe2+ → 2Cr3++ 7H2O + 6Fe3+

(ii) No. of moles of Cr2O72– = cv/1000

= (0.0180 × 19.6)/1000 = 3.528 × 10–4 moles

Molesofiron(II)=3.528×10–4 × 6 = 2.1168 × 10–3 = 0.002118

moles FeSo4·7H2O = moles Fe2+ = 2.1168 × 10–3 mol

M(FeSo4·7H2O) = 277.9 g mol–1

∴ mass FeSo4·7H2O = 2.1168 × 10–3 × 277.9 = 0.588 g

∴ % purity = 0.588 ______ 0.655

× 100 = 89.8%

2 (a) (i) Blue to green to yellow solution. (ii) Thelonepaironthechlorideionisdonatedto

thecopper(II)ion.Aco-ordinate(dativecovalent)bond is formed.

(b) A light blue precipitate forms, which is soluble in excess ammonia forming a deep blue solution.

(c) Ammonia molecule has one lone pair of electrons andthreebondingpairs.Thelonepairrepelsthebonding pairs more than they repel each other, reducing the bond angle. Each ammonia ligand bonded to copper has four bond pairs. So equal repulsion between bonding pairs. Angle larger, at 109.5.

3 Transitionelementsaredblockelementsthathaveanion with an incomplete d sub-shell.

Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9

Note that the d sub-shell is incomplete.

2�

Cu

H2O

H2O OH2

OH2

OH2

OH2

Theshapeistetrahedral. Water molecules are ligands (electron-pair donors) and

theseformco-ordinatebondstothecentralmetal.TheCu2+ ion is an electron-pair acceptor.

• Transitionmetalshavevariableoxidationstates,forexample Cu2+ and Cu+.

• Transitionmetalsformcolouredcompounds:CuCl42– is yellow. [Cu(H2O)6]2+ is blue.

• Transitionmetalsandtheirionsactascatalysts.Ironis a catalyst for the Haber process.

4 (a) 1s2 2s2 2p6 3s2 3p6 3d5

Ironhasanincompletedsub-shellintheFe3+ion.Itis a transition metal.

(b) 1 Variable oxidation state. 2 Forms coloured compounds. (c) Fe2+ reacts with hydroxide ions to form a green

precipitate whereas Fe3+ reacts with hydroxide to form an orange-brown precipitate.

(d) 4Fe2+ + O2 + 4H+ → 4Fe3+ + 2H2O (e) (i) Copper may react with MnO4

–. Copper reacts with Fe3+ reducing it back to Fe2+. (ii) Mn in MnO4

– changes from oxidation state +7 to +2 in Mn2+ (it is reduced). Fe changes from Fe2+ (+2) to Fe3+ (+3) and is oxidised.


(iii) No. of moles of MnO4– = (22.5 × 0.02)/1000

= 4.50 × 10–4

Moles of Fe2+ = 5 × 4.50 × 10–4 = 2.25 × 10–3

Mass of Fe2+ = 2.25 × 10–3 × 55.8 = 0.1256 g Percentage = (0.1256/0.675) × 100 = 18.6% 5 (a) A pale blue solution forms a light blue precipitate

whichissolubleinexcessammoniatogiveadeepblue solution.

(b)

2�

Cu

H2O

H2O OH2

OH2

OH2

OH2

(c) Waterhas2lonepairsand2bondingpairs.Thetwo lone pairs repel the bonding pairs more than the bond pairs repel each other. Bond angle is 104.5º but in the ligand a lone pair is used to form a co-ordinate bond to the metal. Water ligand has 3 bondpairsandonelonepair.Thereislessrepulsion. So the bond angle is greater than in a water molecule.

6 (a) Transitionelementsarecatalysts. (b) (i)

Cl

C

Cl

ClCl

109.5°2�

(ii) Cl–

(iii) Concentrated hydrochloric acid. (iv) [Cu(H2O)6]2+ + 4Cl– → CuCl42– + 6H2O Thisisareactionwhereoneligandisreplaced

by another. Here H2O is replaced by Cl–. 7 (a) Oxygen changes from –1 (H2O2) to –2 (in H2O): it is

reduced. Oxygen changes from –1 (H2O2) to 0 (O2): it is

oxidised. (b) (i) 2MnO4

– + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2

(ii) No. of moles of MnO4– = (17.5 × 0.02)/1000

= 0.00035 No. of moles of H2O2 = (0.00035/2) × 5

= 0.000875 0.000875 = (concentration × 25)/1000 Concentration = 0.0350 mol dm–3

(c) Name of reagent used: NaOH Observation:Red–brownprecipitate.

8 (a) (i) +3 (ii)

�

Ru

H2O

Cl OH2

Cl

OH2

OH2

Trans

�

Ru

H2O

Cl OH2

OH2

Cl

OH2

Cis

(iii) Cis–trans. (b) (i) HCl(conc). (ii) Ligand substitution. 9 (a) (i) Thisisthenumberofco-ordinatebondsformed

by ligands to the transition metal ion. (ii) Square planar. (b) (i) Ligand substitution. (ii) x = 2– y = 0 (c) (i)

PtCl

Cl NH3

NH3

PtH3N

Cl NH3

Cl

O O

(ii) Cis–trans. (iii) Cis-platinbindswithDNAandpreventscell

division,stoppingthegrowthoftumours.10 (a) (i) [Fe(NH3)4Cl2]+

(ii)

Cis Trans

FeCl

H3N NH3

Cl

NH3

NH3

CrCl

H3N Cl

NH3

NH3

NH3��

(iii) 6 (b) Cisplatinisananti-cancerdrug.ItbindstoDNA

andpreventsreplication,stoppingthegrowthoftumours.

11 (a) +2 (b) No. of moles of S2O3

2– = (0.100 × 22.0)/1000 Answer: 000220 mol (c) No.ofmolesofI2 = 0.00220/2 = 0.0011 Answer: 000110 mol (d) No. of moles of Cu2+ = 0.00110 × 2 Answer: 000220 mol (e) 0.00220 = (concentration × 25.0)/1000 Answer: 0.0880 mol dm–3


12 (a) (i) From pink to blue (ii) Tetrahedral (iii) Ligand substitution (b)

N

H

C

H

H

C

H

H

N

H

HH

Co

NH2

NH2

NH2

NH2

H2N

H2NCo

H2N

H2N

H2N

NH2

NH2

NH2

3� 3� (c) (i) Optical isomerism (ii)

N

H

C

H

H

C

H

H

N

H

HH

Co

NH2

NH2

NH2

NH2

H2N

H2NCo

H2N

H2N

H2N

NH2

NH2

NH2

3� 3�

13 (a) 1s2 2s2 2p6 3s2 3p6 3d6; Fe2+ has an incompletely filled d subshell.

(b) (i) [Fe(H2O)6]2+ (ii)

2�

Fe

H2O

H2O OH290°

OH2

OH2

OH2

(iii) Each ligand donates an electron pair to the metaliontoformacoordinate(dativecovalent)bond.

(c) [Fe(H2O)6]3+ + 4Cl– Y [FeCl4]2+ + 6H2O (d) FeCl2givesagreenprecipitateandFeCl3givesa

brown-red precipitate. Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) (e) Cr changes from +3 to +6, which is oxidation; Fe changes from +6 to +3, which is reduction. (f) 4FeO4

2– + 20H+ → 4Fe3+ + 3O2 + 10H2O

14 (a) MoO3 + 2Al → Mo + Al2O3

(b) [Kr]4d3, which has an incompletely filled d sub-shell, meaning it is a transition metal ion.

(c) Cr2O72–(aq) + 2H+(aq) + 3MoO2(s) →

3MoO42–(aq) + 2Cr3+(aq) + H2O(l)

(d) (i) K2FeO4

(ii) Moles of KOH = 4.00 × 10.0/1000 = 0.0400 mol Moles of Fe2O3 = 1.00/159.6 = 0.00627 mol From the equation, 1 mol Fe2O3 reacts with

10 mol OH– ≡ 10 mol KOH ∴ 0.00400 mol Fe2O3 reacts with 0.0400 mole

KOH. ∴ Fe2O3 is in excess by 0.00227 mol.

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