Unit 2 Equilibria, energetics and elements (F325 ...UNIT+2+Module+3... · ... energetics and...

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Unit 2 Equilibria, energetics and elements

(F325) Transition elements

Topics covered in this module:

1. Transition metals

2. Properties of transition metal compounds

3. Catalysis and precipitation

4. Transition metals and complex ions

5. Stereoisomerism in complex ions

6. Bidentate and multidentate ligands

7. Ligand substitution in complexes

8. Ligand substitution and stability constants

9. Redox titrations

10. Examples of redox titrations

11. Redox titrations – iodine and thiosulfate

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1. Transition metals

Unit 2 – Equilibria, energetics and elements

Transition elements Done

Deduce the electron configurations of atoms and ions of the d-block elements.

Describe the elements Ti-Cu as transition elements.

Key areas to concentrate on…

You must be able to deduce the electron configurations of the transition metals and their ions.

You may be required to work out electron configurations of unusual ions based upon this knowledge.

Remember that the 4s sub-shell has a lower energy than the 3d sub-shell. This means that the 4s orbital fills before the orbitals in the 3d sub-shell. The d-block metals will have filled their 4s orbital before electrons are added to the 3d sub-shell.

When writing the electron configurations, the official way is to show all the n = 3 sub-shells before the n = 4 energy level. In your textbook, the energy levels have been shown in order of sub-shell filling with the 4s before the 3d. Either response is perfectly acceptable in exams, so cobalt can be shown as:

1s22s22p63s23p64s23d7 or 1s22s22p63s23p63d74s2

When forming ions in transition element chemistry, the 4s electrons are always removed before any electrons are taken from the 3d sub-shells.

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Past paper questions

Mercury forms two ions, Hg22+ and Hg2+. The table shows the electronic

configuration of mercury in these ions.

ion electronic configuration

Hg22+ [Xe]4f145d106s1

Hg2+ [Xe]4f145d10

Use the electronic configurations to explain why mercury is not a transition element.

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[Total 1 mark]

Examiner’s comments

This question involved the chemistry of mercury and hydrogen peroxide.

Almost all candidates were able to apply the information about the electronic configuration of mercury.

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Iron forms several complex ions in which the oxidation state of iron is +3.

(i) Complete the electronic configuration for an iron(Ill) ion, Fe3+.

1s22s22p6 ...................................................................................................

[1]

(ii) Explain, using electronic configuration, why iron is a transition element.

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[1]

[Total 2 marks]


Candidates found this question about iron(III) one of the more accessible questions on the examination paper.

Only a small proportion of the candidates could not give the electronic configuration for an iron(III) ion and an even smaller proportion could not explain why iron is a transition element.

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2. Properties of transition metal compounds



Illustrate the existence of more than one oxidation state for a transition element in its compounds.

Illustrate the formation of coloured metal ions.


You must know the four key properties of transition elements:

good catalysts

variable oxidation states

formation of complex ions

formation of coloured ions.

It is always a good idea to be able to quote an example for each property.

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An unusual compound of iron has been detected on the surface of the planet Mars. This compound contains the ferrate(VI) ion.

A student uses 1.00 g of iron(III) oxide and makes, on crystallisation, 0.450 g of sodium ferrate(VI), Na2FeO4.

Calculate the percentage yield, by mass, of sodium ferrate(VI).

Show your working.

Express your answer to an appropriate number of significant figures.

percentage yield = ............................ %

[Total 4 marks]


This question focussed on the ferrate(VI) ion and involved synoptic assessment of topics such as redox and calculations.

Candidates of all ability were able to make progress and this question was rarely left blank. Good answers were well laid out and it was very easy to follow the steps involved. Candidates had to quote the answer to three significant figures to get full marks and full credit was given to either 21.6. or 21.7.

Common misconceptions included the wrong formula to calculate the relative formula

masses e.g. using FeO42– instead of Na2FeO4. Other candidates used the relative atomic

mass of iron instead of the relative formula mass of iron(III) oxide. A significant proportion of candidates did not take into account the molar ratio of 2 : 1 (Na2FeO4 : Fe2O3).

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Ferrate(VI) ions will decompose in acidic solution as shown in the equation below.

4FeO42–(aq) + 2OH+(aq) → 4Fe3+(aq) + 3O2(g) + 10H2O(l)

Explain, in terms of oxidation numbers, why this decomposition involves both reduction and oxidation.

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[Total 2 marks]



A large proportion of candidates scored at least one mark for calculating the oxidation numbers. The most common misconception was that the oxidation number of oxygen changed from –8 to 0.

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Copper and zinc are both d-block elements but only copper is a transition

element. Copper forms compounds containing Cu2+ or Cu+ ions but zinc only

forms compounds containing Zn2+ ions.

(a) Use the electronic configurations of Cu2+ and Zn2+ to explain why copper is a transition element and zinc is not.

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[2]

(b) Suggest two differences between compounds containing Zn2+ and Cu2+ ions.

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[2]

[Total 4 marks]


This was a short question about brass, copper and zinc.

(a) A smaller proportion of candidates than in previous examination papers wrote incorrect electronic configurations for transition metal ions. Almost without exception candidates could explain why copper is a transition element and zinc is not.

(b) The mark scheme demanded a comment about compounds of Cu2+ and a

comparison with compounds of Zn2+. It was not sufficient just to state that copper salts are coloured. Candidates made comparisons about the colour, possible catalytic properties and complex formation. Since the specification makes no reference to the

chemistry of Zn2+ candidates were allowed to state that this ion would not form

complexes. A common misconception was to refer to Cu2+ having varying oxidation states.

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Brass is an alloy of copper and zinc.

The percentage of copper and zinc in a sample of brass can be determined by reaction with hydrochloric acid. Only zinc reacts, as shown in the equation below.

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g)

• A sample of brass powder of known mass is added to an excess of 1.00

mol dm–3 hydrochloric acid. • The mixture is heated gently and the hydrogen collected is measured

once the reaction has finished.

A student analyses a 1.23 g sample of brass using the method described.

The student collects 76.0 cm3 of hydrogen at room temperature and pressure.

1 mol of gas molecules occupies 24.0 dm3 at room temperature and pressure.

Calculate the percentage by mass of copper in the sample of brass. Give your answer to an appropriate number of significant figures.

answer .............................................. %

[Total 3 marks]


This was a short question about brass, copper and zinc. Candidates found the calculation demanding.

Many candidates were able to calculate the number of moles of hydrogen produced although a frequent misconception was to half this value to get the moles of zinc. A significant proportion of the candidates correctly worked out the mass of zinc but a small number of candidates used the relative atomic mass for copper instead of that for zinc. Only the most able candidates realised that they had to subtract this mass from the total mass to get the mass of copper. The mark scheme gave full marks for answers between 82.9 and 83.2 provided they had three or four significant figures. The majority of candidates quoted answers to the correct number of significant figures.

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Artists between the 13th and the 19th Centuries used a green pigment called verdigris. The artists made the pigment by hanging copper foil over boiling vinegar.

During the preparation of verdigris, copper atoms are oxidised to copper(II) ions.

(i) Write the oxidation half equation for the conversion of copper atoms into copper(II) ions.

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[1]

(ii) The reduction half equation that takes place is as follows.

O2(g) + 4H+(aq) + 4e– 2H2O(l)

Construct the equation for the redox reaction between copper, oxygen and hydrogen ions.

[1]

[Total 2 marks]


This question proved to be quite demanding and many candidates only scored one or two marks out of the five available.

In (i) most candidates wrote the correct half equation. The mark scheme allowed an

equilibrium sign, Cu → Cu2+ + 2e–, or Cu – 2e– → Cu2+.

The most common misconception was to put the electrons on the left hand side

i.e. Cu + 2e– → Cu2+.

In (ii), the full redox equation was often correct but there were often errors of omission, e.g. having only one Cu rather than two.

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3. Catalysis and precipitation



Illustrate the catalytic behaviour of the transition elements and/or their compounds.

Describe the simple precipitation reactions of Cu2+(aq), Co2+

(aq), Fe2+

(aq) and Fe3+(aq) with aqueous sodium hydroxide.


You must also know the four key properties of transition elements:

good catalysts

variable oxidation states

formation of complex ions

formation of coloured ions.

It is always a good idea to be able to quote an example for each property.

You need to describe about the formation and colour of the hydroxide precipitates of:

Fe2+

Fe3+

Co2+

Cu2+

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Excess aqueous sodium hydroxide is added to an aqueous solution containing

Fe3+(aq).

(i) Describe what you would see happen.

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[1]

(ii) Write an ionic equation, including state symbols, for the reaction that takes place.

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[2]

[Total 3 marks]



In (i), many candidates gave the correct colour for iron(III) hydroxide but failed to mention it was a precipitate and were not awarded a mark.

Although a significant proportion of candidates could write the ionic equation they did not always give the correct state symbols; in particular Fe(OH)3(aq) was very common. Only a

very small proportion of the candidates wrote molecular equations but a greater proportion of candidates included the spectator ions.

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The compound FeSO4.7H2O can be used to kill moss in grass. Iron(II) ions in a

solution of FeSO4.7H2O are slowly oxidised to form iron(III) ions.

Describe a test to show the presence of iron(III) ions in a solution of FeSO4.7H2O.

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[Total 1 mark]


Many candidates scored less than half-marks in this question.

Many candidates could give a chemical test for aqueous iron(II) ions. The use of either aqueous sodium hydroxide or aqueous thiocyanate ions was allowed on the mark scheme. A common misconception was that potassium thiosulfate reacts with aqueous iron(II) to make a blood red coloration.

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A moss killer contains iron(II) sulphate.

Some of the iron(II) sulphate gets oxidised to form iron(III) sulphate. During the

oxidation iron(II) ions, Fe2+, react with oxygen, O2, and hydrogen ions to make

water and iron(III) ions, Fe3+.

(a) Complete the electronic configuration for Fe3+ and use it to explain why iron is a transition element.

Fe3+:1s22s22p6 ...........................................................................................

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[2]

(b) State two typical properties of compounds of a transition element.

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[2]

(c) Describe how aqueous sodium hydroxide can be used to distinguish between aqueous iron(II) sulphate and aqueous iron(III) sulphate.

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[2]

(d) Construct the equation for the oxidation of acidified iron(II) ions by oxygen.

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[2]

[Total 8 marks]

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This question was generally the most well answered of the four on the examination paper.

(a) Many candidates were able to give the electronic configuration for the iron(III) ion, however there was a significant proportion of the candidates who added three extra electrons while others included the 4s electrons. Almost all candidates could explain why iron was a transition element.

(b) Most candidates scored full marks for this part question, however a small proportion gave properties of the elements such as strong, hard, or a high melting point; these properties were not given credit.

(c) In this part question the majority of candidates could describe the colour changes that occurred but many did not state that a precipitate or an insoluble solid was made and so were not given credit.

(d) This part question proved to be the most difficult part question on the paper and candidates rarely scored a mark. The most common misconception was to try and write equations that made iron(III) oxide even though the question asked for iron(III) ions.

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4. Transition metals and complex ions



Explain the term ligand and its role in coordinate bonding.

Describe and use the terms complex ion and coordination number.

State and give examples of complexes with six-fold coordination with an octahedral shape.


Ligands play an important part in transition element chemistry. You need to describe:

how ligands bond to ions

the shapes of the complexes formed

the isomerism (cis/trans and optical) shown.

Ligand substitution is also important and you will be required to:

calculate and use the stability constants, Kstab, of complexes

know the colour changes associated with ligand substitution listed in the specification.

In exams you are expected to be able to draw a three-dimensional diagram for a complex ion. Transition metal complex ions are drawn using wedge-shaped bonds, which indicate bonds coming out of the plane of the paper. Hatched lines show bonds going into the paper. Solid lines are in the plane of the paper.

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A sample of iron is heated with a stream of dry hydrogen chloride. A different

chloride of iron is formed that contains the Fe2+ ion. This chloride dissolves in water to form a pale green solution that contains the hexaaquairon(II) complex ion.

(i) Complete the electronic configuration of Fe2+.

1s22s22p6 ...................................................................................................

[1]

(ii) Draw the shape of the hexaaquairon(II) complex ion. Include the bond angles on your diagram.

[2]

(iii) Aqueous sodium hydroxide is added to a solution containing Fe2+(aq).

State what you would observe.

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Write an ionic equation, with state symbols, for the reaction.

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[2]

[Total 5 marks]

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This question was focussed on the chemistry of iron(II) and iron(III). It combined aspects of synoptic assessment and knowledge and understanding from the Trends and Patterns part of the specification. A greater proportion of the candidates obtained high marks in this question than in the previous two questions. A small but significant proportion of the candidates did not attempt the calculations.

(i) Many candidates got the correct answer but a small proportion of candidates gave

1s22s22p63s23p63d44s2.

(ii) The majority of candidates could draw the shape of the octahedral complex, with only a small number drawing square planar or tetrahedral. Fewer candidates gave the correct bond angle; some left it out and others gave the tetrahedral angle.

(iii) The mark scheme in required the formation of a green precipitate or a green solid. Many candidates did not include the precipitate or the solid. In the ionic equation the majority of candidates either ignored state symbols or put Fe(OH)2(aq) rather than

Fe(OH)2(s).

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One of these complex ions is [Fe(CN)6]3–. This is called the hexacyanoferrate(III)

ion. In the hexacyanoferrate(III) ion the cyanide ions, CN–, act as ligands.

(i) Suggest why a cyanide ion can act as a ligand.

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[1]

(ii) Draw the expected shape for the complex ion [Fe(CN)6]3–. Include bond

angles and the name of the shape.

[2]

[Total 3 marks]



Even though cyano complexes are not covered by the specification, a large proportion of candidates could suggest why the cyanide ion could act as a ligand. To gain credit candidates had to mention the presence of a lone pair and that it could be donated. No penalty was incurred by candidates referring to the lone pair on nitrogen being donated.

A very large proportion of candidates could draw a three-dimensional representation of an octahedral shape but a smaller proportion could state the correct bond angle. To gain full marks candidates had to also state the name of the shape as indicated in the question.

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Ammonia is a simple molecule. The H—N—H bond angle in an isolated ammonia molecule is 107°.

The diagram shows part of the [Cu(NH3)4(H2O)2]2+ ion and the H—N—H bond

angle in the ammonia ligand.

Explain why the H—N—H bond angle in the ammonia ligand is 109.5° rather than 107°.

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[Total 3 marks]


Candidates often were imprecise when referring to ammonia – it was not always clear whether it was the ligand or the isolated molecule. A large proportion of candidates realised that lone pairs were more repelling but often candidates then went on to state that bonds or hydrogen atoms were repelled.

H

Cu2+

N 109.5º

H

H

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Cobalt readily forms complex ions in which the cobalt has an oxidation state of +2.

One complex ion of cobalt is the hexaaquocobalt(II) ion [Co(H2O)6]2+.

(i) What is the co-ordination number of Co2+ in this complex ion?

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[1]

(ii) Water is acting as a ligand. Explain the meaning of the term ligand.

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[2]

[Total 3 marks]


(i) Candidates are well aware of the concept of co-ordination number. A few confused this with oxidation number.

(ii) There were many full descriptions of the term ligand.

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Water is a simple molecule. The H—O—H bond angle in an isolated water molecule is 104.5°.

The diagram shows part of the [Cu(H2O)6]2+ ion and the H—O—H bond angle

in the water ligand.

Explain why the H—O—H bond angle in the water ligand is 107° rather than 104.5°.

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[Total 3 marks]


Although similar questions have been asked in previous examination papers many candidates did not refer to electron pair repulsion. A significant number of candidates referred to repelling atoms and bonds. Only a small proportion of the candidates stated specifically that an isolated water molecule has two lone pairs and as a ligand it has only one lone pair. The last marking point related to the difference in repelling power of bond pairs and lone pairs.

Cu +2 O

H

H

107º

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Transition metals readily form complex ions when they are combined with a suitable ligand.

What is meant by the following terms?

(i) complex ion

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[1]

(ii) ligand

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[2]

[Total 3 marks]


Both parts of this question were well answered.

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In this question, one mark is available for the quality of use and organisation of scientific terms.

Copper and iron are typical transition elements. One of the characteristic properties of a transition element is that it can form complex ions.

• Explain in terms of electronic configuration why copper is a transition

element.

• Give an example of a complex ion that contains copper. Draw the three dimensional shape of the ion and describe the bonding within this complex ion.

• Transition elements show typical metallic properties. Describe three other typical properties of transition elements. Illustrate each property using copper or iron or their compounds.

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[Turn over]

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[11]

Quality of Written Communication [1]

[Total 12 marks]

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Many candidates answered this question about copper complexes and the properties of transition elements very well. A significant proportion of the candidates obtained full or nearly full marks for this question.

A small proportion of the candidates did not score the mark available for the quality of written communication in this question. Candidates needed to use at least three of the technical terms listed in the mark scheme in the correct context.

The mark scheme had two marks allocated to why copper is a transition element, six marks for the copper complex and three marks to the properties of the transition elements.

Most candidates were able to explain why copper is a transition element but a significant proportion of candidates did not write or could not write the electronic configuration of the copper(II) ion. Many candidates that did write the electronic configuration included 4s electrons.

Most candidates gave either [Cu(H2O)6]2+ or [CuCl4]2– as examples of complex ions. The

most popular complex ion given was [Cu(H2O)6]2+. Some candidates forgot to include the

charge on the complex ions. The bonding within this complex ion was well known, with many candidates giving both a written explanation and a diagram. The most likely omission was not to state that the central metal ion is an electron pair acceptor.

Although most candidates could give the correct name for the shape of the complex ion they often gave the wrong bond angle. Diagrams of the octahedral complex should indicate the three-dimensional shape with the use of wedges, and candidates that did not were not credited.

Only catalytic properties, coloured compounds or ions and several oxidation states were given credit for typical properties of transition elements or their compounds. To be awarded a mark each property had to have an associated example. The most popular example for a catalyst was iron in the Haber process. Often incorrect oxidation numbers for copper were included and so the mark was not awarded. A small proportion of the candidates gave physical properties typical of all metals and these were not given credit.

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5. Stereoisomerism in complex ions



Describe stereoisomerism in transition metal complexes using examples of cis-trans and optcal isomerism.

Describe the use of cis-plain as an anti-cancer drug, and its action by binding to DNA.









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In this question, one mark is available for the quality of use and organisation of scientific terms.

Stereoisomerism is very common in transition metal complexes. Some complexes have found an important use in the treatment of cancer.

(i) Name a transition metal complex used in the treatment of cancer.

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[1]

(ii) Describe how this complex helps in the treatment of cancer.

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[2]

[Total 3 marks]


Most candidates are aware of cis-platin and its use in the treatment of cancer, but only stronger candidates scored all 3 marks by explaining how it bonds to DNA and prevents replication.

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6. Bidentate and multidentate ligands



Explain and use the term bidentate ligand (e.g. NH2CH2CH2NH2, ‘en’).

Describe stereoisomerism in transition element multidentate complexes using examples of cis-trans and optical isomerism.









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The ethanedioate ion, C2O42–, can act as a bidentate ligand when it forms

complex ions with a transition metal ion. The structure of the ethanedioate ion is shown below.

(a) What do you understand by the term bidentate ligand?

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[2]

(b) The ethanedioate ion readily forms an octahedral complex ion with Cr3+.

Show the structure and charge of this complex ion.

[3]

[Total 5 marks]


(a) Many candidates scored both marks here but too many lost a mark by omitting to mention either the two ‘lone pairs’ or the coordinate bonds formed.

(b) Good candidates had little trouble in drawing a 3D diagram with three ethanedioate ions bonded correctly and with the correct charge. Some candidates did not show the ethanedioate ions bonding via the single bonded oxygens and many did not give a correct overall charge.

O

C

–O

O

O–

C

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(a) Co2+ forms the complex [Co(NH3)4Cl2]. This complex exists as two

stereoisomers.

(i) Draw diagrams to show the two isomeric forms of this complex.

[2]

(ii) What type of stereoisomerism is shown by this complex?

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[1]

(b) Cobalt also forms a complex with the formula [Co(H2NCH2CH2NH2)2Cl2].

This complex shows the same kind of isomerism as [Co(NH3)4Cl2] but it

also shows a different type of stereoisomerism.

Draw diagrams to show the two isomers of this different type of stereoisomerism.

[2]

[Total 5 marks]


(a) (i) The majority of candidates provided 3-D diagrams.

(ii) Most candidates recognised this as cis-trans isomerism.

(b) A common mistake was to draw the [Co(en)3]2+ complex. Some candidates chose

the trans isomer instead of the cis isomer as being optically active.

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The edta4– ion forms complex ions with Ni2+(aq).

(a) Complete the electronic configuration of the Ni2+ ion.

Is22s22p6 ....................................................................................................

[1]

(b) The edta4– ion has the following structure.

(i) Put a ring around two different types of atom in the edta4– ion that

are capable of forming a dative covalent bond with the Ni2+ ion.

[2]

(ii) What feature of these atoms allows them to form a bond with Ni2+?

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[1]

[Total 4 marks]


(a) A surprising number of candidates included 4s2 in the electron configuration of the

Ni2+ ion.

(b) The nitrogen atom was often not recognised as an atom capable of forming a dative covalent bond.

O

C

O

C

O

CH2

N

CH2

CH2 CH2 N

O O

C

O

O

C

O

H C2

H C2

–

–

–

–

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(a) A common ligand which combines with a number of transition metal ions is ethane-1,2-diamine, H2NCH2CH2NH2. This is a bidentate ligand.

Explain the meaning of the term bidentate.

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[1]

(b) The complex [CoCl2(H2NCH2CH2NH2)2] is a neutral molecule. It shows

two types of stereoisomerism. Use this molecule to explain what you understand by the term stereoisomerism. Your answer should include diagrams to show clearly the structures of the different isomers in both types of stereoisomerism.

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[7]

[Total 8 marks]

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(a) Most candidates understood the term ‘bidentate’ but a few gave imprecise answers by stating ‘more than one’ lone pair of electrons.

(b) Few candidates scored full marks on this part. There was some confusion about stereoisomerism in that many candidates thought that the isomers had a different structure. Many candidates did not refer to optical isomerism at all or chose the trans isomer before drawing the mirror images. It was common to see ‘optical isomers are non superimposable’ without the words ‘mirror images’.

A significant number of candidates drew square planar complexes with some ingenious arrangements of ethanediamine as a monodentate ligand. As mentioned in the general comments, there are still candidates drawing non 3-D diagrams to show the arrangement of bonds in complexes.

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Describe the types of stereoisomerism found in transition metal complexes.

Use suitable examples to illustrate your answer.

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[8]

Quality of Written Communication [1]

[Total 9 marks]

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More able candidates had an excellent opportunity to score high marks on this part and they provided well-organised answers with good examples, illustrated well with diagrams of the 3-dimensional nature of transition metal complexes.

Weaker candidates scored marks for cis-trans isomerism but were often less confident with optical isomerism.

Many candidates showed the ethanediammine ligand as having terminal –NH3 groups and

this was penalised once. Weak candidates have difficulty drawing this complex in three dimensions and the second isomer is then not always a mirror image. Candidates should be advised to always draw a mirror line so that the position of the mirror image is clear.

A majority of candidates were awarded the mark for Quality of Written Communication by using at least 3 technical words in an appropriate way.

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7. Ligand substitution in complexes



Describe the process of ligand substitution.

Describe examples of ligand substitution using [Cu(H2O)6]2+ and

[Co(H2O)6]2+ with ammonia and chloride ions.









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Ruthenium (Ru) is a metal in the second transition series. It forms complex ions with the following formulae.

A = [Ru(H2O)6]3+

B = [Ru(H2O)5Cl]2+

C = [Ru(H2O)4Cl2]+

(a) (i) What is the oxidation number of ruthenium in B?

oxidation number of ruthenium = .........................................................

[1]

(ii) One of the complex ions, A, B or C, shows stereoisomerism.

Draw diagrams to show the structures of the two isomers.

[2]

(iii) Name this type of stereoisomerism.

...........................................................................................................

[1]

(b) The complex ion [Ru(H2O)6]3+ can be converted into [Ru(H2O)5Cl]2+.

(i) Suggest a suitable reagent for this conversion.

...........................................................................................................

[1]

(ii) What type of reaction is this?

...........................................................................................................

[1]

[Total 6 marks]

39 | P a g e


(a) (i) This was an easy start to the paper but quite a few candidates did not respond with +3.

(ii) Most candidates correctly chose C. 3-D diagrams are now being used in all but a few cases.

(iii) There was some confusion here but a majority responded with the correct type of stereoisomerism, cis-trans or geometric.

(b) (i) This was generally well-answered but Cl– was not accepted as a reagent.

(ii) Most candidates recognised this as ligand substitution.

40 | P a g e


Dilute aqueous copper(II) sulphate contains [Cu(H2O)6]2+ ions.

(a) Concentrated hydrochloric acid is added drop by drop to a small volume of dilute aqueous copper(II) sulphate. The equation for the reaction taking place is as follows.

[Cu(H2O)6]2+(aq) + 4Cl–(aq) [CuCl4]2–(aq) + 6H2O(l)

(i) Describe the observations that would be made during the addition of the concentrated hydrochloric acid.

...........................................................................................................

[1]

(ii) Describe the bonding within the complex ion, [CuCl4]2–.

...........................................................................................................

...........................................................................................................

...........................................................................................................

[2]

(b) Concentrated aqueous ammonia is added drop by drop to aqueous copper(II) sulphate until present in excess. Two reactions take place, one

after the other, to produce the complex ion [Cu(NH3)4(H2O)2]2+(aq).

Describe the observations that would be made during the addition of concentrated aqueous ammonia.

....................................................................................................................

....................................................................................................................

[2]

[Total 5 marks]

41 | P a g e


Candidates found part (b) much more difficult that the recall questions in part (a).

(a) A large proportion of the candidates were able to give the correct colour change in part (i); both yellow and green were allowed on the mark scheme.

In part (ii), many candidates scored full marks by referring to the ligand having a lone pair that is donated to the copper ion forming a dative bond.

(b) Only a very small proportion of the candidates could recall the observations when aqueous ammonia is added to aqueous copper(II) ions. Many candidates did not state whether a solution or a precipitate was present. Typically answers stated that the mixture went light blue and then dark blue, this was not given any credit in the mark scheme.

42 | P a g e


Cobalt readily forms complex ions in which the cobalt has an oxidation state of +2.

[Co(H2O)6]2+ readily undergoes the following reaction.

[Co(H2O)6]2+(aq) + 4Cl–(aq) [CoCl4]2–(aq) + 6H2O(l)

(i) What is the shape of each complex in this reaction?

[Co(H2O)6]2+ shape ......................................................................................

[CoCl4]2– shape ......................................................................................

[1]

(ii) What colour change would occur on going from left to right in this reaction?

from ................................................... to ....................................................

[1]

(iii) What type of reaction is taking place when [Co(H2O)6]2+ reacts with Cl–?

....................................................................................................................

[1]

[Total 3 marks]


(i) The shapes of complexes are well known.

(ii) The colour change was sometimes reversed, and was also confused with that associated with the copper(II) complex.

(iii) A few candidates did not use the word ligand when describing this substitution reaction.

43 | P a g e


Platinum forms complexes with a co-ordination number of 4.

(a) (i) Explain the term co-ordination number.

...........................................................................................................

...........................................................................................................

[1]

(ii) State the shape of these platinum complexes.

...........................................................................................................

[1]

(b) The tetrachloroplatinate(II) ion readily undergoes the following reaction.

[PtCl4]x + 2NH3 [Pt(NH3)2Cl2]y + 2Cl–

(i) What type of reaction is this?

...........................................................................................................

[1]

(ii) Suggest values for x and y in the equation.

x = ……………

y = ……………

[2]

(c) The complex [Pt(NH3)2Cl2]y exists in two isomeric forms.

(i) Draw diagrams to show the structure of these isomers.

[2]

[Turn over]

44 | P a g e

(ii) What type of isomerism is this?

...........................................................................................................

[1]

(iii) One of the isomers of [Pt(NH3)2Cl2]y is an important drug used in

the treatment of cancer.

How does this drug help in the treatment of cancer?

...........................................................................................................

...........................................................................................................

...........................................................................................................

...........................................................................................................

[2]

[Total 10 marks]


(a) Many candidates described, incorrectly, that co-ordination number is the number of ligands attached, rather than the number of coordinate bonds. Many gave the shape of platinum complexes as tetrahedral.

(b) There were some very interesting values for x and y.

(c) Many candidates drew tetrahedral structures here but then used the structure to describe cis-trans isomerism. The use of cis-platin as a treatment for cancer is well known.

45 | P a g e


Aqueous copper(II) sulphate contains [Cu(H2O)6]2+ ions. Aqueous ammonia is

added drop by drop to a small volume of aqueous copper(II) sulphate. Two reactions take place, one after the other, as shown in the equations.

[Cu(H2O)6]2+(aq) + 2OH–(aq) Cu(OH)2(s) + 6H2O(l)

Cu(OH)2(s) + 2H2O(l) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 2OH–

(aq)

(a) Describe the observations that would be made as ammonia is added drop by drop until it is in an excess.

....................................................................................................................

....................................................................................................................

[2]

(b) Draw the shape for the [Cu(H2O)6]2+ ion. Include the bond angles in your

diagram.

[2]

[Total 4 marks]


Many candidates answered this question about copper complexes very poorly.

(a) In this part, candidates often described the colour changes without stating whether there was a solution or a precipitate formed. The state symbols given in the equations in the stem were not always used by candidates to deduce the observations.

(b) In this part, many candidates drew an octahedral complex showing the three dimensional nature using wedges and/or dotted line. Drawings not indicating three dimensions were not given credit. Bond angles between 60° and 120° were often quoted. The best answers gave more than one correct bond angle typically both 90° and 180°. Some candidates even showed the bond angles within the water ligands.

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The Co2+ ion can form complexes with two different co-ordination numbers.

(a) What is meant by the co-ordination number of a complex ion?

....................................................................................................................

....................................................................................................................

[1]

(b) The following equilibrium is readily established.

[Co(H2O)6]2+ + 4Cl– [CoCl4]2– + 6H2O

(i) In the boxes below, draw the shape of each complex ion.

[Co(H2O)6]2+ [CoCl4]2–

[2]

(ii) What colour change would you expect to see when an excess of

Cl– is added to [Co(H2O)6]2+?

from .............................................. to ...............................................

[2]

(iii) Describe how you would move the position of this equilibrium to the left.

...........................................................................................................

[1]

[Total 6 marks]

47 | P a g e


(a) Many candidates defined co-ordination number as the number of ligands rather than the number of co-ordinate bonds attached to the central metal ion.

(b) (i) Diagrams must show the 3-dimensional arrangement of bonds. Many

candidates showed [CoCl4]2– as square planar rather than tetrahedral.

(ii) Many candidates did not know this colour change. Some used the colours associated with the change from chromate(VI) to dichromate(VI), whilst others simply got the colours the wrong way round. Most candidates were familiar with

Le Chatelier but many suggested adding more [CoCl4]2– to move the position of

equilibrium to the left. Some candidates suggested increasing the concentration of water, which was not credited.

48 | P a g e

8. Ligand substitution and stability constants



Explain the biochemical importance of iron in haemoglobin, including ligand substitution.

Use and understand the term stability constant, Kstab.

Deduce expressions for Kstab of ligand substitutions and understand that a large Kstab results in formation of a stable complex ion.









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9. Redox titrations



Describe redox behaviour in transition elements using suitable examples.

Carry out redox titrations and carry out structured calculations involving MnO4

-.


You may encounter redox titrations – these are similar to the acid base titrations you met in AS Unit F321: Module 1 - Atoms and reactions. However, unlike at AS:

the ratios of reactants you encounter at A2 may be in the order of 5:1 or 2:5 – very different from the 1:1 ratio or 1:2 ratio seen at AS!

you will be expected to carry out the calculations without any guidance from the exam paper.

50 | P a g e



Aqueous sodium ferrate(VI) is a very powerful oxidising agent.

(i) Predict what you would see when aqueous sodium ferrate(VI) is added to aqueous potassium iodide.

Explain your answer.

....................................................................................................................

....................................................................................................................

[1]

(ii) Aqueous sodium ferrate(VI) will oxidise ammonia into substance X.

Suggest an identity for X.

....................................................................................................................

[1]

[Total 2 marks]



In part (i), the mark scheme required that the species responsible for the observation was also noted down. Although a significant proportion referred to a brown colour being produced often a reference to iodine was not made.

Although a significant proportion of candidates gave correct answers to part (ii), including oxides of nitrogen, elemental nitrogen and hydroxylamine others gave answers such as

NH2–, NH3O or NH4

+.

51 | P a g e


Vanadium can exist in a number of different oxidation states. One compound

of vanadium is ammonium vanadate(V) and this contains the ion VO3–. This

can be reduced to V2+ in several steps, using zinc metal and aqueous

sulphuric acid.

(a) 25.0 cm3 of 0.100 mol dm–3 ammonium vanadate(V) is completely

reduced to V2+(aq) using zinc and aqueous sulphuric acid. The resulting

solution is titrated with 0.0500 mol dm–3 MnO4–(aq) and 30.0 cm3 is

required to oxidise the V2+(aq) back to VO3–(aq).

The half equation for acidified MnO4– acting as an oxidising agent is

shown below.

MnO4– + 8H+ + 5e– Mn2+ + 4H2O

Show that the vanadium has changed oxidation state from +2 to +5 in this titration.

[4]

(b) Suggest an equation for the oxidation of V2+(aq) to VO3–(aq) by MnO4

–

(aq) under acid conditions.

....................................................................................................................

[2]

[Total 6 marks]

52 | P a g e


(a) Despite being given volumetric data, many candidates answered this question using oxidation numbers only. Such answers were not credited. Only the best candidates obtained full marks for this part. Weaker candidates were able to work out the number of

moles of MnO4– and VO3

– but could not then make the link between these values and the

number of electrons transferred.

(b) Only the most able candidates in selected centres were able to provide a correct,

balanced equation for the oxidation of V2+ to VO3–. This would suggest that the skill of

constructing redox equations from first principles is not prioritised in a many centres.

53 | P a g e

10. Examples of redox titrations



Understand how to carry out redox titrations and carry out structured calculations involving MnO4

- and I2/S2O32-.

Perform non-structured titration calculations, based on experimental results.





54 | P a g e


Mercury thermometers are not used in some laboratories because of the danger of mercury vapour. This vapour is very easily absorbed through the lungs into the blood.

In the blood, mercury reacts with hydrogen peroxide to form mercury(II) oxide.

Hg + H2O2 → HgO + H2O

The mercury(II) oxide formed accumulates within organs in the body.

Use oxidation numbers to show that the reaction between mercury and hydrogen peroxide is an example of both oxidation and reduction.

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

[Total 2 marks]


This question involved the chemistry of mercury and hydrogen peroxide.

A large proportion of candidates was able to get at least one mark. This was usually for identifying that mercury is being oxidised, together with the correct change in oxidation number. Many candidates did not understand that the oxidation state of oxygen in hydrogen peroxide was –1 and so had hydrogen being reduced.

55 | P a g e


The percentage purity of a sample of manganese(IV) oxide, MnO2, can be

determined by its reaction with acidified iron(II) ions.

• Stage 1 – A sample of known mass of the impure MnO2 is added to a conical

flask.

• Stage 2 – The sample is reacted with a known excess amount of Fe2+ acidified with dilute sulphuric acid.

• Stage 3 – The contents of the flask are heated gently.

• Stage 4 – The cooled contents of the flask are titrated with aqueous potassium manganate(VII) in acidic conditions to find the amount

of unreacted Fe2+.

(i) The reduction half-equation for manganese(IV) oxide in the presence of dilute acid is shown below.

MnO2(s) + 4H+(aq) + 2e– → Mn2+(aq) + 2H2O(l)

Construct the balanced equation for the redox reaction between

Fe2+(aq), MnO2(s) and H+(aq).

....................................................................................................................

....................................................................................................................

....................................................................................................................

[1]

(ii) In Stage 1 and Stage 2 a student uses a 0.504 g sample of impure MnO2

and 100 cm3 of 0.200 mol dm–3 Fe2+.

In Stage 4 the student determines that the amount of unreacted Fe2+ is 0.0123 mol.

1 mol of MnO2 reacts with 2 mol of Fe2+.

Calculate the percentage purity of the impure sample of MnO2.

percentage purity = ..................................................... %

[3]

[Total 4 marks]

56 | P a g e


This question was focussed on the chemistry of iron(II) and iron(III). It combined aspects of synoptic assessment and knowledge and understanding from the Trends and Patterns part of the specification. A greater proportion of the candidates obtained high marks in this question than in the previous two questions. A small but significant proportion of the candidates did not attempt the calculations.

(i) Many candidates wrote the correct ionic equation but a small proportion of

candidates wrote equations involving Fe and Fe2+ or the reduction of Fe3+.

(ii) Only a small proportion of candidates were able to calculate the percentage purity. Answers between 66.3 and 66.5 were allowed on the mark scheme.

Two approaches were used by candidates.

Approach 1

• Calculate amount of Fe2+ that reacted = 0.0077 mol

• Calculate mass of MnO2 that reacted = 0.335 g

• Determine percentage purity

Approach 2

• Calculate amount of MnO2 if sample was pure = 0.00580 mol

• Calculate amount of Fe2+ that would have reacted with this pure sample = 0.0116 mol

• Calculate amount of Fe2+ that actually reacted = 0.0077 mol

• Determine percentage

The majority of candidates failed to get even the first steps correct.

57 | P a g e



When chlorine is bubbled through a suspension of iron(III) oxide in concentrated aqueous sodium hydroxide, a solution of aqueous sodium ferrate(VI) forms.

The two relevant redox systems are shown below.

Cl2(aq) + 2e– → 2C/–(aq)

Fe2O3(s) + 10OH–(aq) → 2FeO42–(aq) + 5H2O(I) + 6e–

Construct the redox equation for the reaction between chlorine, iron(III) oxide and hydroxide ions.

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

[Total 2 marks]



The majority of candidates were able to construct a redox equation. One mark was given for equations that still included electrons or contained the correct reactants and products but were not balanced.

58 | P a g e


The compound FeSO4.7H2O can be used to kill moss in grass.

The percentage purity of an impure sample of FeSO4.7H2O can be determined

by titration against potassium dichromate(VI), K2Cr2O7, under acid conditions,

using a suitable indicator.

During the titration, Fe2+(aq) ions are oxidised to Fe3+(aq) ions.

• Stage 1 – A sample of known mass of the impure FeSO4.7H2O is added

to a conical flask.

• Stage 2 – The sample is dissolved in an excess of dilute sulphuric acid.

• Stage 3 – The contents of the flask are titrated against K2Cr2O7(aq).

(i) The reduction half equation for acidified dichromate(VI) ions, Cr2O72–, is

as follows.

Cr2O72–(aq) + 14H+(aq) + 6e– 2Cr3+(aq) + 7H2O(l)

Construct the balanced equation for the redox reaction between

Fe2+(aq), Cr2O72–(aq) and H+(aq).

....................................................................................................................

....................................................................................................................

....................................................................................................................

[2]

(ii) In Stage 1, a student uses a 0.655 g sample of impure FeSO4.7H2O.

In Stage 3, the student uses 19.6 cm3 of 0.0180 mol dm–3 Cr2O72– to

reach the end-point.

One mole of Cr2O72– reacts with 6 moles of Fe2+.

Calculate the percentage purity of the impure sample of FeSO4.7H2O.

percentage purity .........................

[4]

[Total 6 marks]

59 | P a g e


In part (i) a significant proportion of the candidates successfully constructed the balanced

equation. Only a small number of candidates did not realise that Fe2+ was oxidised to give

Fe3+.

In part (ii) only a small proportion of candidates were able to calculate the percentage purity as 89.9 (89.6 to 90.0 were accepted on the mark scheme). A larger proportion

scored three marks; typically these used the mass of Fe2+ rather than of FeSO4.7H2O.

Candidates used a variety of different approaches to this calculation, e.g.

• calculating moles of Cr2O72–, then moles of Fe2+ in sample, then mass of

FeSO4.7H2O in sample and finally the percentage

• calculating moles of Cr2O72–, then moles of Fe2+ in sample then moles of

FeSO4.7H2O and finally the percentage

• calculating moles of FeSO4.7H2O in 0.655 g, then moles of Cr2O72– needed to react

with this, then moles of Cr2O72– in titre and finally the percentage.

60 | P a g e

11. Redox titrations – iodine and thiosulfate



Carry out structural calculations involving I2/S2O32-.

Perform unstructured titration calculations based on experimental results.





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The standard electrode potential of Cu2+(aq) + 2e– Cu(s) is +0.34 V.

A student measured the standard electrode potential of Cu2+(aq) + 2e– Cu(s). She was surprised to see that the emf of the cell was less than the expected value of +0.34 V.

She decided to measure the concentration of the Cu2+(aq) ions in the solution by titration.

25.00 cm3 of the solution containing Cu2+ ions were pipetted into a volumetric

flask and she made the volume up to 250.0 cm3 with distilled water.

An excess of aqueous potassium iodide, KI, was added to 25.00 cm3 of the diluted solution.

The iodine formed was titrated against 0.100 mol dm–3 sodium thiosulfate, Na2S2O3.

The volume of Na2S2O3(aq) used was 23.20 cm3.

The equations for the formation and titration of iodine are given below.

2Cu2+(aq) + 4I–(aq) → 2CuI(s) + I2(aq)

I2(aq) + 2S2O32–(aq) → 2I–(aq) + S4O6

2–(aq)

(i) State how the student would identify the end point of the titration.

....................................................................................................................

....................................................................................................................

[1]

(ii) Show that the concentration of the Cu2+(aq) ions was 0.93 mol dm–3.

[4]

[Turn over]

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(iii) Explain, in terms of chemical equilibrium, why the emf of this cell was less than the standard electrode potential.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[2]

[Total 7 marks]


(i) This was very disappointing. Few candidates seemed to know that the end point for this titration was an off-white precipitate. The common answer was brown to colourless or blue to colourless if starch was used.

(ii) This caused real problems for a number of candidates. Even good candidates often lost a mark by omitting to scale down by a factor of 10 to account for dilution. It was

common to see the correct answer of 0.93 mol dm–3 written down from a calculation that

gave the answer of 0.093 mol dm–3.

(iii) Candidates needed to state that 0.93 mol dm–3 or a concentration less than 1.00

mol dm–3 or standard to obtain a mark here before proceeding to explain how this would

affect the position of equilibrium and the value of Eo.

63 | P a g e


Chromium metal and its compounds have a number of important uses.

Cr2O72– ions oxidise I– ions to I2 under acid conditions according to the

following equation.

Cr2O72–(aq) + 6I–(aq) + 14H+(aq) 2Cr3+(aq) + 3I2(aq) + 7H2O(l)

(i) If you carried out this reaction, how could you see that iodine is formed?

....................................................................................................................

....................................................................................................................

[1]

(ii) How could you use the formation of I2 in this reaction to determine the

concentration of a solution of Cr2O72– ions?

In your answer

• state the method you would use

• state the reagents used

• show how you would use your results.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[4]

[Total 5 marks]


(i) Most candidates gave the colour of iodine as brown.

(ii) There were many disappointing answers to this part. Whilst a majority of candidates realised that a titration should be carried out, very few suggested that sodium thiosulfate should be used and hardly anyone gave an equation for the reaction of iodine

and the thiosulfate ion or determined the molar relationship between Cr2O72– and S2O3

2–.

64 | P a g e


Potassium dichromate(VI) can be used in a number of redox reactions. The standard electrode potentials for two half reactions are given below.

Cr2O72– + 14H+ + 6e–

2Cr3+ + 7H2O Eο = +1.33 V

I2 + 2e– 2I– Eο = +0.54 V

Acidified potassium dichromate(VI) is added to aqueous potassium iodide to give aqueous iodine.

(i) Construct an ionic equation to show the reaction taking place when acidified potassium dichromate(VI) is added to aqueous potassium iodide.

....................................................................................................................

....................................................................................................................

....................................................................................................................

[2]

(ii) An excess of aqueous sodium thiosulfate was then added. Describe and explain what you would see.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[3]

[Total 5 marks]


Many candidates were able to combine the half equations given to construct the overall redox equation. A few candidates managed to have iodine on the LHS, i.e. as a reagent rather than a product.

Part (ii) proved to be quite challenging. Many candidates failed to link sodium thiosulfate to iodine titrations. Some thought that thiosulfate and thiocyanate were the same species. Better candidates, who did realise that thiosulfate would decolorise iodine, often

did not realise that the final colour would be green due to Cr3+ ions.

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Brass is a widely used alloy of copper. It is possible to analyse a sample of brass by initially dissolving it in concentrated nitric acid.

(a) (i) What other metal is present in brass?

...........................................................................................................

[1]

(ii) Give one common use for brass and state the property of brass which makes it ideal for that purpose.

...........................................................................................................

...........................................................................................................

[1]

(b) During the analysis of brass, 1.65 g of the alloy was reacted with concentrated nitric acid. The resulting solution was neutralised,

transferred to a volumetric flask and made up to 250 cm3 using distilled water.

An excess of aqueous potassium iodide was added to a 25.0 cm3 portion of the solution from the volumetric flask and the liberated iodine was

titrated with 0.100 mol dm–3 sodium thiosulfate. 20.0 cm3 of aqueous sodium thiosulfate were required to remove the iodine.

(i) What could be used to neutralise the excess nitric acid?

...........................................................................................................

[1]

(ii) What indicator is used in the titration of iodine with sodium thiosulfate?

...........................................................................................................

[1]

(iii) When is this indicator added to the titration mixture?

...........................................................................................................

[1]

[Turn over]

66 | P a g e

(c) The reactions taking place in this titration may be summarised as follows.

2Cu2+ + 4I– 2CuI + I2

I2 + 2S2O32–

2I– + S4O62–

(i) Calculate the amount, in moles, of sodium thiosulfate in 20.0 cm3 of solution.

answer ......................... mol

[1]

(ii) For every one mole of Cu2+ ions present in solution, deduce the

amount, in moles, of S2O32– ions needed for the titration.

answer ......................... mol

[1]

(iii) What is the amount, in moles, of Cu2+ ions present in 25.00 cm3 of solution?

answer ......................... mol

[1]

(iv) Calculate the percentage by mass of copper in the sample of brass.

answer ...................... % Cu

[3]

[Total 11 marks]

67 | P a g e


(a)

(i) Nickel and tin were often incorrectly suggested as the second metal in brass.

(ii) Most candidates could give a use of brass but did not give the property of brass that made it useful for that purpose.

(b)

(i) Many candidates could name an alkali here. The word ‘alkali ‘on its own was not credited.

(ii) Surprisingly few candidates used starch as the indicator with many suggesting methyl orange or phenolphthalein. These candidates appeared to be unfamiliar with iodine/thiosulfate titrations.

(iii) As with (b)(ii), the unfamiliarity of these titrations was obvious in many candidates.

(c)

Good candidates did not have a problem with the volumetric calculations and there were many high scores for this part. One common mistake in (iii) was to use the 25 cm3 of solution to obtain the wrong answer of 0.0025 mol. Another error was in (iv), when candidates often failed to scale up to 250 cm3 and ending with 7.7% for the amount of copper in brass. These candidates were not awarded the final mark.

68 | P a g e

Specimen paper questions

This question looks at the chemistry of transition elements.

(a) (i) Explain what is meant by the terms transition element, complex ion and ligand,

(ii) Discuss, with examples, equations and observations, the typical reactions of transition elements.

In your answer you should make clear how any observations provide evidence for the type of reaction discussed.

[11]

(b) Describe, using suitable examples and diagrams, the different shapes and stereoisomerism shown by complex ions.

In your answer you should make clear how your diagrams illustrate the type of stereoisomerism involved.

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[9]

[Total 20 marks]

71 | P a g e

Sample student answers

A2 Unit F325: Equilibria, energetics and elements

Module 6: Transition elements

Question 1

Total marks: 15

(a) Iron is a typical transition element. (i) What is meant by the term transition element? (ii) Write the electron configuration for a Fe

2+ ion.

(iii) Using iron as your example, state the typical properties of a transition element. (iv) Describe what you would see when Fe

3+(aq) ions react with aqueous sodium

hydroxide and write the ionic equation.

Marks available: (i) 1 (ii) 1 (iii) 4 (iv) 2

Student answer:

(a) (i) An element which can form an ion with an incomplete d sub-shell. (ii) 1s

22s

22p

63s

23p

63d

6

(iii) Using iron:

Form coloured compounds, e.g. Fe2+

ions are green Form complex ions, e.g. Fe(H2O)6

2+

Have ions with different oxidation states, e.g. Fe2+

is +2 and Fe3+

is +3 It shows good catalytic properties, e.g. Fe in the Haber process. (iv) An orange/brown precipitate would form. Fe

3+(aq) + 3OH

–(aq) Fe(OH)3(s)

Examiner comments:

(a) (i) The student has correctly stated an ion with an incomplete d sub-shell (would also allow half-filled d-orbital). (ii) A common error is to remove two 3d electrons and leave the two 4s electrons, instead of removing the two 4s electrons. (iii) Iron is a good example to use as it is relatively easy to show how it exhibits all four of the typical properties of a transition element. (iv) Although not asked for here, you should include state symbols when writing ionic equations.

72 | P a g e

(b) What is meant by the term bidentate ligand?

Marks available: 2

Student answer:

(b) A ligand is a species with a lone pair of electrons with which it can form a coordinate bond to a transition metal ion. Bidentate means the ligand species has two lone pairs of electrons with which it can form a coordinate bond to a transition metal ion.

Examiner comments:

(b) This is an example of where there are two words in italics –you need to address both terms.

(c) Draw a diagram to show and explain the isomerism exhibited by [Ni(NH2CH2CH2NH2)3]2+

.

Marks available: 2

Student answer:

(c)

Examiner comments:

(c) The diagrams have correctly been drawn to represent 3D drawings.

73 | P a g e

(d) State the shape of the following complex ions. (i) CuCl4

2–

(ii) Ni(NH3)2Cl2

(iii) Fe(H2O)62+

Marks available: 3

Student answer:

(d) (i) Tetrahedral (ii) Square planar (iii) Octahedral

Examiner comments:

(d) The three shapes given here are the three most common, so are likely to feature widely.

Module 6: Transition elements

Question 2

Total marks: 15

(a) A student added concentrated hydrochloric acid to a solution of aqueous cobalt chloride. (i) Describe what the student would observe. (ii) Write the equation for the reaction which occurred and state what type of process happened during the reaction.

Marks available: (i) 1 (ii) 3

Student answer:

(a) (i) The pink solution turned blue. (ii) Co(H2O)6

2+(aq) + 4Cl

–(aq) CoCl4

2–(aq) + 6H2O(l)

The process is called ligand exchange.

Examiner comments:

(a) (i) It’s a good idea to include the initial colour as well as the final colour. (ii) Although the state symbols are not required, you do need to know the formulae of the complexes, including the charges. The equation must also be balanced.

74 | P a g e

(b) Many aqueous transition element ions undergo reactions with cyanide CN–(aq) ions,

according to the equation shown below: Fe(H2O)6

2+(aq) + 6CN

–(aq) Fe(CN)6

y(aq) + 6H2O(l)

(i) Calculate the value of y. (ii) Write the equation for the stability constant Kstab for this reaction, giving its units.

Marks available: (i) 1 (ii) 2

Student answer:

(b) (i) y = 4-.

(ii) Kstab = [Fe(CN)64–

](aq) units dm18

mol–6

[Fe(H2O)6

2+(aq)] + [CN

–(aq)]

5

Examiner comments:

(b) (i) The student has realised that the Fe2+

contributes +2 in terms of oxidation number and the 6CN

– ions contributes 6 × -1 in terms of oxidation number, giving an overall charge

of –4 on the left-hand side of the equation. So, y must be 4– to balance the equation in

terms of charge.

(ii) Note how the student has calculated the units. Each [ ] means mol dm–3

so:

on the bottom line the units are {mol dm–3

+ (6 × mol dm–3

)} giving mol7 dm

–21 in

total

on the top line the units are mol dm–3

in total

top units divided by bottom units gives dm18

mol–6

.

75 | P a g e

(c) A student carried out a series of steps to determine the percentage of copper in brass.

Step 1 0.724 g of brass was added to excess nitric acid. Step 2 Excess aqueous potassium iodide was added to this mixture.

Step 3 The excess iodine formed was titrated against 0.400 mol dm–3

aqueous sodium

thiosulfate and an end-point of 24.35 cm3 was recorded as the colour of iodine

disappeared to leave a white precipitate. (i) Write the equation for the reaction that occurred in Step 2. (ii) Write the equation for the reaction that occurred in Step 3. (iii) Calculate the percentage of copper in this sample of brass.

Marks available: (i) 1 (ii) 1 (iii) 4

Student answer:

(c) (i) 2Cu2+

(aq) + 4I– (aq) 2CuI(s) + I2(aq)

(ii) I2(aq) + 2S2O3

2–(aq) 2I

– (aq) + S4O6

2–(aq)

(iii) Moles of S2O32–

(aq) in equation (ii) = 0.400 × 24.35 = 9.74 × 10–3

mol 1000

Moles of Cu2+

(aq) in equation (i) = 9.74 × 10–3

mol

Mass of Cu metal = 9.74 × 10–3

mol × 63.5 = 0.618 g (3 sf) (0.61849) % Cu present in sample = 0.618 × 100 = 85.4% 0.724

Examiner comments:

(c) (i) The student has correctly identified that copper(I) iodide is formed and has balanced the equation accordingly. (ii) The S4O6

2–(aq) ion is unusual but the equation has been balanced correctly.

(iii) Although all three sub-answers are shown rounded to three significant figures (3 sf), it is acceptable to leave your sub-answers in the calculator’s memory and round off your final answer.

76 | P a g e

(d) Copper(I) sulfate is a white solid. When added to water, the white solid dissolves and a blue solution with some brown particles in the bottom remains. Explain the redox processes which occurred during the washing process.

Marks available: 2

Student answer:

(d) The copper has changed from a +1 oxidation state in copper(I) sulfate to Cu2+

ions (which have a +2 oxidation state) and solid Cu metal (which has a 0 oxidation state). This process is called disproportionation.

Examiner comments:

(d) Remember to write in terms of oxidation numbers and not ionic charge.

77 | P a g e

The Periodic Table of the Elements

Th

e P

eri

od

ic T

ab

le o

f th

e E

lem

en

ts

1

2

3

4

5

6

7

0

Ke

y

1.0

H

h

yd

roge

n

1

4.0

H

e

heliu

m

2

6.9

L

i lit

hiu

m

3

9.0

B

e

bery

lliu

m

4

rela

tive a

tom

ic m

ass

ato

mic

sym

bo

l n

am

e

ato

mic

(pro

ton)

num

ber

10.8

B

b

oro

n

5

12.0

C

ca

rbo

n

6

14.0

N

nitro

gen

7

16.0

O

oxygen

8

19.0

F

fluo

rin

e

9

20.2

N

e

neon

10

23.0

N

a

sodiu

m

11

24.3

M

g

ma

gnesiu

m

12

27

.0

Al

alu

min

ium

13

28.1

S

isili

co

n

14

31.0

P

p

hosp

horu

s

15

32.1

S

sulfur

16

35.5

C

l ch

lorine

17

39.9

A

rarg

on

18

39.1

K

pota

ssiu

m

19

40.1

C

a

calc

ium

20

45.0

S

c

sca

nd

ium

21

47.9

T

i

tita

niu

m

22

50.9

V

vana

diu

m

23

52.0

C

rchro

miu

m

24

54

.9

Mn

ma

ngane

se

25

55

.8

Fe

iro

n

26

58.9

C

ocoba

lt

27

58.7

N

inic

kel

28

63

.5

Cu

co

ppe

r

29

65.4

Z

n

zin

c

30

69

.7

Ga

ga

llium

31

72.6

G

ege

rma

niu

m

32

74.9

A

sars

enic

33

79.0

S

esele

niu

m

34

79.9

B

rb

rom

ine

35

83.8

K

rkry

pto

n

36

85.5

R

b

rubid

ium

37

87.6

S

r str

ontium

38

88.9

Y

ytt

rium

39

91.2

Z

r

zirco

niu

m

40

92.9

N

b

nio

biu

m

41

95.9

M

om

oly

bdenum

42

[98]

Tc

tech

ne

tiu

m

43

101

.1

Ru

ru

the

niu

m

44

102

.9

Rh

rhod

ium

45

10

6.4

P

d

palla

diu

m

46

10

7.9

A

g

silv

er

47

112.4

C

d

cad

miu

m

48

114

.8

Inin

diu

m

49

118

.7

Sn

tin

50

121.8

S

ban

tim

on

y

51

127

.6

Te

tellu

rium

52

126

.9

I io

din

e

53

131.3

X

exen

on

54

13

2.9

C

s

cae

siu

m

55

137

.3

Ba

b

arium

56

138

.9

La*

lan

thanum

57

178.5

H

f

ha

fniu

m

72

18

0.9

Ta

tan

talu

m

73

18

3.8

W

tu

ngste

n

74

186.2

R

e

rheniu

m

75

190

.2

Os

osm

ium

76

192

.2

Ir

irid

ium

77

19

5.1

P

t

pla

tinum

78

19

7.0

A

u

gold

79

20

0.6

H

g

merc

ury

80

20

4.4

T

l th

alli

um

81

20

7.2

P

ble

ad

82

209.0

B

ib

ism

uth

83

[209

] P

op

olo

niu

m

84

[210

] A

ta

sta

tine

85

[222

] R

nra

do

n

86

[22

3]

Fr

franciu

m

87

[226

] R

a

rad

ium

88

[227

] A

c*

actiniu

m

89

[261]

Rf

ruth

erf

ord

ium

104

[262]

Db

d

ubniu

m

105

[26

6]

Sg

seab

org

ium

10

6

[264]

Bh

b

oh

rium

107

[27

7]

Hs

ha

ssiu

m

10

8

[26

8]

Mt

me

itne

rium

10

9

[271]

Ds

darm

sta

dtium

11

0

[272

] R

g

roen

tge

niu

m

111

Ele

men

ts w

ith

ato

mic

nu

mbe

rs 1

12–

116 h

ave b

een

rep

ort

ed b

ut n

ot fu

lly

au

then

ticate

d

140.1

C

e

ce

rium

58

14

0.9

P

r pra

seo

dym

ium

59

14

4.2

N

dne

odym

ium

60

144.9

P

mpro

meth

ium

61

150

.4

Sm

sam

arium

62

152

.0

Eu

euro

piu

m

63

15

7.2

G

dga

do

liniu

m

64

15

8.9

T

b

terb

ium

65

16

2.5

D

ydysp

rosiu

m

66

16

4.9

H

oh

olm

ium

67

16

7.3

E

re

rbiu

m

68

168.9

T

mth

uliu

m

69

173

.0

Yb

ytt

erb

ium

70

175

.0

Lu

lute

tium

71

232.0

T

h

tho

rium

90

[231]

Pa

pro

tactiniu

m

91

23

8.1

U

ura

niu

m

92

[237]

Np

nep

tun

ium

93

[24

2]

Pu

plu

toniu

m

94

[24

3]

Am

am

ericiu

m

95

[247]

Cm

curi

um

96

[245

] B

k

be

rkeliu

m

97

[251]

Cf

ca

liforn

ium

98

[25

4]

Es

ein

ste

iniu

m

99

[253]

Fm

ferm

ium

100

[256

] M

dm

ende

levium

101

[254

] N

on

ob

eliu

m

102

[257

] L

rla

wre

nciu

m

103

• • • • • •

Unit 2 Equilibria, energetics and elements (F325 ...UNIT+2+Module+3... · ... energetics and...

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Transcript of Unit 2 Equilibria, energetics and elements (F325 ...UNIT+2+Module+3... · ... energetics and...