O C R A2 C H E M I S T R Y MODULE 5 - Exam QA

O C R A 2 C H E M I S T R Y

MODULE 5.6

1

TRANSITION ELEMENTS

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In the Contact Process sulfur dioxide reacts with oxygen to form sulfur trioxide as shown in theequation.

2SO2(g) + O2(g) ⇌ 2SO3(g)

The table shows some thermodynamic data.

∆HfΘ/kJ mol–1 SΘ/J K–1 mol–1

SO2(g) –297 248

O2(g) 0 205

SO3(g) –395 256

(a) Use data from the table to calculate the standard enthalpy change for this reaction.

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(b) Use data from the table to calculate the standard entropy change for this reaction.

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(c) State what the sign of the entropy change in your answer to part (b) indicates about theproduct of this reaction relative to the reactants.

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(d) Use your answers to parts (a) and (b) to calculate a value for the free–energy change forthis reaction at 50°C.

(If you were unable to calculate ∆H in part (a) assume a value of –250 kJ mol–1.If you were unable to calculate ∆S in part (b) assume a value of –250 J K–1 mol–1.These are not the correct values.)

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(e) Use your answer to part (d) to explain whether the reaction is feasible at 50°C.

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(f) Vanadium(V) oxide acts as a heterogeneous catalyst in the Contact Process.

(i) State what is meant by the term heterogeneous.

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(ii) Write two equations that show how this catalyst is involved in the Contact Process.

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(iii) Suggest why the vanadium(V) oxide is used in small pellet form rather than as largelumps.

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(iv) State why the reactants should be purified before they come into contact with thevanadium(V) oxide.

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(Total 14 marks)


This question is about cobalt chemistry.

(a) Give the electron configuration of the Co atom and of the Co2+ ion.

State three characteristic features of the chemistry of cobalt and its compounds.

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(b) Ethane-1,2-diamine can act as a bidentate ligand. When [Co(H2O)6]2+(aq) ions are treatedwith an excess of ethane-1,2-diamine, the water ligands are replaced.

Explain what is meant by the term bidentate ligand.

Explain, with the aid of an equation, the thermodynamic reasons why this reaction occurs.

Draw a diagram to show the structure of the complex ion formed.

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(Total 12 marks)


A student weighed out a 2.29 g sample of impure K3[Fe(C2O4)3].3H2O and dissolved it in water.

This solution was added to a 250 cm3 volumetric flask and made up to 250 cm3 with distilledwater.A 25.0 cm3 portion was pipetted into a conical flask and an excess of acid was added.The mixture was heated to 60°C and titrated with 0.0200 mol dm–3 KMnO4 solution.

26.40 cm3 of KMnO4 solution were needed for a complete reaction.

In this titration only the C2O42– ions react with the KMnO4 solution.

(a) The reaction between C2O42– ions and MnO4– ions is autocatalysed.

Explain what is meant by the term autocatalysed and identify the catalyst in the reaction.

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(b) Select from the list the most suitable substance used to acidify the solution in the conicalflask.Put a tick (✔) in the correct box.

H2C2O4

H2SO4

HCl

HNO3

(1)


(c) The reaction between C2O42– ions and MnO4

– ions is very slow at first.Explain why the reaction is initially slow.

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(d) Write an equation for the reaction between C2O42– ions and MnO4

– ions in acidic solution.Calculate the percentage purity of the original sample of K3[Fe(C2O4)3].3H2O.Give your answer to 3 significant figures.

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(e) A solution of KMnO4 has an unknown concentration.

Describe briefly how colorimetry can be used to determine the concentration of thissolution.

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(Total 16 marks)

(a) Explain how the electron pair repulsion theory can be used to deduce the shape of, and thebond angle in, PF3

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(b) State the full electron configuration of a cobalt(II) ion.

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(c) Suggest one reason why electron pair repulsion theory cannot be used to predict theshape of the [CoCl4]2− ion.

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(d) Predict the shape of, and the bond angle in, the complex rhodium ion [RhCl4]2−.

Shape ............................................................................................................

Bond angle .....................................................................................................(2)

(Total 10 marks)

The table below shows some successive ionisation energy data for atoms of three differentelements X, Y and Z.

Elements X, Y and Z are Ca, Sc and V but not in that order.

First Second Third Fourth Fifth Sixth

X 648 1370 2870 4600 6280 12 400

Y 590 1150 4940 6480 8120 10 496

Z 632 1240 2390 7110 8870 10 720

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(a) Which element is calcium?

X

Y

Z (1)

(b) Which element is vanadium?

X

Y

Z (1)

(c) Justify your choice of vanadium in part (b)

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(d) An acidified solution of NH4VO3 reacts with zinc.

Explain how observations from this reaction show that vanadium exists in at least twodifferent oxidation states.

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(e) The vanadium in 50.0 cm3 of a 0.800 mol dm−3 solution of NH4VO3 reacts with 506 cm3 ofsulfur(IV) oxide gas measured at 20.0 °C and 98.0 kPa.

Use this information to calculate the oxidation state of the vanadium in the solution after thereduction reaction with sulfur(IV) oxide.Explain your working.The gas constant R = 8.31 J K−1 mol−1.

Oxidation state = ...............................(6)

(Total 11 marks)


(a) A co-ordinate bond is formed when a transition metal ion reacts with a ligand.

Explain how this co-ordinate bond is formed.

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(b) Describe what you would observe when dilute aqueous ammonia is added dropwise, toexcess, to an aqueous solution containing copper(II) ions.Write equations for the reactions that occur.

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(c) When the complex ion [Cu(NH3)4(H2O)2]2+ reacts with 1,2-diaminoethane, the ammoniamolecules but not the water molecules are replaced.

Write an equation for this reaction.

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(d) Suggest why the enthalpy change for the reaction in part (c) is approximately zero.

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(e) Explain why the reaction in part (c) occurs despite having an enthalpy change that isapproximately zero.

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(Total 11 marks)

A green solution, X, is thought to contain [Fe(H2O)6]2+ ions.

(a) The presence of these ions can be confirmed by reacting separate samples of solution Xwith aqueous ammonia and with aqueous sodium carbonate.

Write equations for each of these reactions and describe what you would observe.

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(b) A 50.0 cm3 sample of solution X was added to 50 cm3 of dilute sulfuric acid and made up to250 cm3 of solution in a volumetric flask.

A 25.0 cm3 sample of this solution from the volumetric flask was titrated with a 0.0205 moldm−3 solution of KMnO4

At the end point of the reaction, the volume of KMnO4 solution added was 18.70 cm3.

(i) State the colour change that occurs at the end point of this titration and give a reasonfor the colour change.

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(ii) Write an equation for the reaction between iron(II) ions and manganate(VII) ions.

Use this equation and the information given to calculate the concentration of iron(II)ions in the original solution X.

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(Total 11 marks)


The redox reaction, in aqueous solution, between acidified potassium manganate(VII) andsodium ethanedioate is autocatalysed.

(a) Write an equation for this redox reaction.

Identify the species that acts as the catalyst.

Explain how the properties of the species enable it to act as a catalyst in this reaction.

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(b) Sketch a graph to show how the concentration of MnO4− ions varies with time in this

reaction.Explain the shape of the graph.

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(Total 10 marks)


Chlorine can be found in water. One method for the determination of chlorine in water is to usecolorimetry.

A colourless sample of water from a vase of flowers was analysed after the addition of compoundZ as the addition of Z resulted in a purple solution.

Compound W

(a) Calculate the Mr of Compound W.

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(b) Determine the percentage, by mass, of nitrogen in this compound.

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(c) A simplified diagram of a colorimeter is shown below.

(i) Suggest why it is important that the container for each sample has the samedimensions.

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(ii) Suggest why the coloured filter is used.

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(iii) Suggest one reason why a colorimetric method might be chosen in preference totitration.

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(Total 5 marks)

Consider the following reaction scheme that starts from aqueous [Cu(H2O)6]2+ ions.

green-blue precipitate

For each of the reactions 1 to 4, identify a suitable reagent, give the formula of the copper-containing species formed and write an equation for the reaction.

(a) Reaction 1

Reagent ........................................................................................................

Copper-containing species ...........................................................................

Equation ........................................................................................................(3)

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(b) Reaction 2

Reagent ........................................................................................................


Equation ........................................................................................................(3)

(c) Reaction 3

Reagent ........................................................................................................


Equation ........................................................................................................(3)


(d) Reaction 4

Reagent ........................................................................................................


Equation ........................................................................................................(3)

(Total 12 marks)


Mark schemes

(a) ∆HΘ = Σ∆ products - Σ∆ reactants

1

or (2 × –395) – (2 × –297)

= –196 (kJ mol–1)

Penalise incorrect units, ignore missing units1

1

(b) ∆SΘ = ΣSΘ products - ΣSΘ reactants1

= (2 × 256) – 205 – (2 × 248)

= –189 JK-1 mol-1

Allow -0.189 kJ K-1 mol-1

Units must be given and must match value1

(c) Causes an increase in order / a decrease in disorder

Allow products more ordered / products less disordered

If answer to (b) is +ve, allow products are less ordered / causes anincrease in disorder / causes a decrease in order

1

(d) ∆GΘ = ∆HΘ - T∆SΘ

Do not insist on standard state symbol1

= –196 – 323 (–189/1000)

If conversion of T or ∆S incorrect, then can only score M11

= –134.9 kJ mol-1

Must have correct units

Allow answers in J mol-1

–135 kJ mol-1

If both alternative values used then -169(.3) kJ mol-1

Allow alternative ∆H and/or alternative ∆S in calculation1

(e) Feasible because ∆G is negativeAllow mark if a correct deduction from answer to (d)

Both a reference to feasibility and to ∆G needed1

(f) (i) (The catalyst is in) a different state or phase (from thereactants)

1


(ii) SO2 + V2O5 → SO3 + V2O4

Allow 2VO2 instead of V2O4

Allow multiples1

Must have equations in this order1

(iii) Surface area is increased1

(iv) So that the catalyst is not poisoned

Allow correct reference to the blocking active sites1

[14]

(a) [Ar] 4s2 3d7 or 1s22s22p63s23p64s23d7

Allow 4s and 3d in either order1

[Ar] 3d7 or 1s22s22p63s23p63d7

1

Any 3

Variable oxidation state

Act as catalysts

Form complexes

Form coloured ions/compounds3

2


(b) Two atoms that each donate a lone pair (of electrons) / coordinatebonds from two atoms

1

Formula of ethane-1,2- diamine: NH2CH2CH2NH2

M2 gained from equation or structure1

[Co (H2O)6]2+ +3NH2CH2CH2NH2 → [Co(NH2CH2CH2NH2)3]2+ +6H2O

Equation must be balanced inc charges

Allow en or C2H8N2 in equation for ethane-1,2-diamine1

There is an increase in the number of particles / the reaction goesfrom 4 moles to 7 moles

Allow increase number of molecules/moles. Allow numbers thatmatch an incorrect equation

1

Disorder/entropy increases / ∆S is positive1

∆G negative1

Mark for correct structure ( ignore charges -even if wrong)


NH2 can be shown in either way – see structure1

[12]

(a) A reaction that produces its own catalyst/ one of the products is the catalyst1

Mn2+

Allow Mn3+

1

3

(b) H2SO41

(c) There is no/very little catalyst at the start OR the reaction onlyspeeds up when the catalyst is produced

1

Two negative ions (MnO4- and C2O4

2-) repel

Reference to molecules loses M21

The activation energy for the reaction is high / heat is required to overcome the activationenergy

1


(d) M1 5 C2O42-(aq) + 2 MnO4

-(aq) + 16 H+(aq) → 10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l)

Ignore state symbols1

M2 n(MnO4–) = OR n(MnO4

-) = 5.28 × 10-4

1

M3 n(C2O42-) = × 5.28 × 10-4 = 1.32 × 10-3

M3 is for M2 × 5/2

If wrong ratio used then can only score M2, M4, M5 and M61

M4 n(C2O42– in flask originally) = 1.32 × 10–3 × 10 = 1.32 × 10-2

M4 is for M3 × 101

M5 n(K3[Fe(C2O4)3].3H2O) = = 4.40 × 10-3

(Mr K3[Fe(C2O4)3].3H2O = 491.1)

M5 is for M4 ÷ 31

M6 Mass of K3[Fe(C2O4)3].3H2O reacted = 4.40 × 10-3 × 491.1 = 2.16 g

M6 is for M5 × 491(.1)1

M7 % purity = x100 = 94.3 or 94.4%

Answer must be to 3 s.f.

Correct answer scores 6 marks; mark equation separately

Alternative method using ratio by moles:

M5 n(C2O42-) = 4.66 × 10-3 × 3 = 0.0140 moles in 250cm3

M6 n(complex) = 2.29/491.1 = 4.66 × 10-3 moles in 250cm3

M7 % = 0.0132/0.0140 × 100 = 94.3 or 94.4%1


(e) Make some known concentrations (of the coloured solution andread the absorbance of each one using a colorimeter)

Ignore addition of suitable ligand1

Plot a graph of absorbance vs concentration

Not just “plot a calibration curve” / reference to Beer-Lambert graphis insufficient

Do not allow transmittance in M21

Read/compare unknown concentration from calibration curve/graph (and hence theconcentration from the graph)

M3 can only be scored if graph/curve mentioned1

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(a) This question is marked using levels of response. Refer to the Mark Scheme Instructionsfor Examiners for guidance on how to mark this question.

All stages are covered and the explanation of each stage is generally correct andvirtually complete.

Answer is communicated coherently and shows a logical progression from stage 1 tostage 2 then stage 3.

Level 35 – 6 marks

4

All stages are covered but the explanation of each stage may be incomplete or maycontain inaccuracies OR two stages are covered and the explanations are generallycorrect and virtually complete.

Answer is mainly coherent and shows progression from stage 1 to stage 3.Level 2

3 – 4 marks

Two stages are covered but the explanation of each stage may be incomplete or maycontain inaccuracies, OR only one stage is covered but the explanation is generallycorrect and virtually complete

Answer includes isolated statements but these are not presented in a logical order orshow confused reasoning.

Level 11 – 2 marks

Insufficient correct chemistry to gain a mark.Level 0

0 marks


Indicative chemistry content

Stage 1: Electrons round P• P has 5 electrons in the outside shell• With 3 electrons from 3 fluorine, there are a total of 8 electrons in outside shell• so 3 bond pairs, 1 non-bond pair

Stage 2: Electron pair repulsion theory• Electron pairs repel as far as possible• Lone pair repels more than bonding pairs

Stage 3: Conclusions• Therefore, tetrahedral / trigonal pyramidal shape• With angle of 109(.5)° decreased to 107°

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(b) 1s22s22p63s23p63d7

Allow correct numbers that are not superscripted1

(c) Too many electrons in d sub-shell / orbitals1

(d) Tetrahedral (shape)1

109.5°

Allow 109°1

[10]

(a) Y15

(b) X1

(c) Jump in trend of ionisation energies after removal of fifth electron

Fits with an element with 5 outer electrons (4s23d3) like V1

(d) Explanation: Two different colours of solution are observed1

Because each colour is due to vanadium in a different oxidation state1

(e) Stage 1: mole calculations in either order

Moles of vanadium = 50.0 × 0.800 / 1000 = 4.00 × 10–2

Extended response

Maximum of 5 marks for answers which do not show a sustainedline of reasoning which is coherent, relevant, substantiated andlogically structured.

1


Moles of SO2 = pV / RT = (98 000 × 506 × 10–6 ) / (8.31 × 293)

= 2.04 × 10–2

1

Stage 2: moles of electrons added to NH4VO3

When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to sulfate(VI) ionsso this is a two electron change

1

Moles of electrons released when SO2 is oxidised = 2.04 × 10–2 × 2

= 4.08 × 10–2

1

Stage 3: conclusion

But in NH4VO3 vanadium is in oxidation state 51

4.00 × 10–2 mol vanadium has gained 4.08 × 10–2 mol of electronstherefore 1 mol vanadium has gained 4.08 × 10–2 / 4.00 × 10 – 2 = 1 molof electrons to the nearest integer, so new oxidation state is 5 – 1 = 4

1[11]

(a) An electron pair on the ligand16

Is donated from the ligand to the central metal ion1

(b) Blue precipitate1

Dissolves to give a dark blue solution1

[Cu(H2O)6]2+ + 2NH3 Cu(H2O)4(OH)2 + 2NH4+

1

Cu(H2O)4(OH)2 + 4NH3 [Cu(NH3)4(H2O)2]2+ + 2OH– + 2H2O1

(c) [Cu(NH3)4(H2O)2]2+ + 2H2NCH2CH2NH2 [Cu(H2NCH2CH2NH2)2(H2O)2]2+ +4NH3

1

(d) Cu–N bonds formed have similar enthalpy / energy to Cu–N bonds broken1

And the same number of bonds broken and made1


(e) 3 particles form 5 particles / disorder increases because more particles are formed /entropy change is positive

1

Therefore, the free-energy change is negative

M2 can only be awarded if M1 is correct1

[11]

(a) [Fe(H2O)6]2+ + 2NH3 → Fe(H2O)4(OH)2 + 2NH4+

Allow equation with OH− provided equation showing formation ofOH− from NH3 given

1

Green precipitate1

[Fe(H2O)6]2+ + CO32− → FeCO3 + 6H2O

1

Green precipitate

effervescence incorrect so loses M41

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(b) (i) Colourless / (pale) green changes to pink / purple (solution)

Do not allow pale pink to purple1

Just after the end−point MnO4− is in excess / present

1

(ii) MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

1

Moles KMnO4 = 18.7 × 0.0205 / 1000 = (3.8335 × 10−4)

Process mark1

Moles Fe2+ = 5 × 3.8335 × 10−4 = 1.91675 × 10−3

Mark for M2 × 51

Moles Fe2+ in 250 cm3 = 10 × 1.91675 × 10−3 = 0.0191675 moles in 50 cm3

Process mark for moles of iron in titration (M3) × 101

Original conc Fe2+ = 0.0191675 × 1000 / 50 = 0.383 mol dm−3

Answer for moles of iron (M4) × 1000 / 50

Answer must be to at least 2 sig. figs. (0.38)1

[11]


(a) 2MnO4− + 16H+ + 5C2O42− → 2Mn2+ + 8H2O + 10CO2

1

Mn2+ OR Mn3+

If catalyst incorrect can only score M1 and M31

(Possible because) Mn can exist in variable oxidation states1

Ea lowered because oppositely charged ions attract

These marks can be gained in any order1

Mn3+ (reduced) to Mn2+ by C2O42− / equation

M5 may appear before M21

Mn2+ (oxidised (back)) to Mn3+ by MnO4− / equation

M5 and M6 can be scored in unbalanced equations or in wordsshowing:

Mn3+ + C2O42− → Mn2+

Mn2+ + MnO4− → Mn3+

1

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(b) Graph marks

S-shaped curve must not risesignificantly and must not fall rapidlyinitially.

Starts on concentration axis and islevelling out (can level out on timeaxis or above but parallel to timeaxis)

Cannot score graph marks (M1 and M2) if no axes and / or nolabels

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Explanation marksSlope / rate increases as catalyst (concentration) forms

1

Slope / rate decreases as (concentration) of MnO4− ions / reactant(s) decreases (OR

reactants are being used up)

Explanation marks can be awarded independent of graph.1

[10]

(a) 164.0

Must be 1 decimal place1

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(b) 17.1(%) (= 28.0 × 100 / Qa)

Consequential on their (a)

Ignore precision but must be to at least 2 sig fig.

(i.e. accept 17 or 17.07)1

(c) (i) Absorption depends on (proportional to) path length / distance travelledthrough solution

Do not allow size.1

(ii) To select the colour / frequency / wavelength that is (most strongly) absorbed(by the sample)

Allow the filter is chosen to complement the colour of the solution1

(iii) Quicker to analyse extracted samples than by titration / uses smaller volumes ofsolution

1[5]

(a) Reaction 1

General principles in marking this question

Square brackets are not essential

Penalise charges on individual ligands rather than on the wholecomplex

Reagent and species can be extracted from the equation

Ignore conditions such as dilute, concentrated, excess

Reagent must be a compound NOT just an ion

Equations must start from [Cu(H2O)6 ]2+ except in part (b)

Mark reagent, species and equation independently

ammonia (NH3) (solution) / NaOH1

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[Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2] + 2NH4+ /

[Cu(H2O)6]2+ + 2OH- → [Cu(H2O)4(OH)2] + 2H2O

Do not allow OH– for reagent

Product 1, balanced equation 1

Allow either equation for ammonia2

(b) Reaction 2

Ammonia (conc / xs)1

[Cu(H2O)4(OH)2] + 4NH3 → [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH−


Note that the equation must start from the hydroxide[Cu(H2O)4(OH)2]

2

(c) Reaction 3

Na2CO3 / any identified soluble carbonate / NaHCO3

Do not allow NaCO3 or any insoluble carbonate but mark on1

[Cu(H2O)6]2+ + CO32- → CuCO3 + 6H2O

OR [Cu(H2O)6]2+ + Na2CO3 → CuCO3 + 6H2O + 2Na+

OR 2[Cu(H2O)6]2+ + 2CO32- → Cu(OH)2.CuCO3 + 11H2O + CO2

OR with NaHCO3

[Cu(H2O)6]2+ + HCO3− → CuCO3 + 6H2O + H+


(d) Reaction 4

HCl (conc / xs) / NaCl

Allow any identified soluble chloride1

[Cu(H2O)6]2+ + 4Cl- → [CuCl4]2- + 6H2O


[12]


O C R A2 C H E M I S T R Y MODULE 5 - Exam QA

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