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Edexcel | Chemistry WCH0 | W 19 5 1 | 1 hour 45 minutes
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20
(a) (I)
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1 Mark

This question is about compounds of copper.

 (a) When copper(II) sulfate solution is mixed with potassium iodide solution,

the copper(II) ions are reduced.  The products are a mixture of iodine and

potassium sulfate in solution, and a white precipitate of a copper compound.

  (i) Write the ionic equation for the reaction of Cu  (aq) and (aq) ions.

   State symbols are not required.

20
(a) (II)
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1 Mark

Complete the electronic configuration for the metal ion in the white precipitate.

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

20
(a) (III)
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1 Mark

 Give the formula of a stable ion of an element in Period 4 of the Periodic Table,

which is isoelectronic with the metal ion in the white precipitate.

20
(b) (I)
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1 Mark

Aqueous copper(II) sulfate contains the complex ion 

State how the electronic energy levels of the copper(II) ion change when 

water ligands form this complex.

 

The (3)d orbitals split / (3)d sub shell splits (into two groups).

ALLOW

(3)d energy level splits

Can be shown on a diagram

20
(b) (II)
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3 Mark s

Explain how this change in energy levels causes complex ions to have

different colours, depending on the ligand.

M1

The gap between groups of energy levels is different with different ligands/

The 3d orbitals split to different extents with different ligands (1)

M2

Electrons absorb/ gain energy of specific frequencies when moving from lower to higher levels

OR

Different frequencies of photons are absorbed when the energy gap differs (1)

M3

The colour seen depends on the energy/ frequency gap (between the two groups of energy levels)

OR

The colour seen is due to the remaining frequencies/ the complementary colour is seen

ALLOW

Colour (seen) is due to reflected light

Colour given out depends on energy gap

(1)

20
(b) (III)
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1 Mark

State the shape of the  ion

Octahedral / octahedron (shape)

IGNORE

diagrams

20
(b) (IV)
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2 Mark s

  ions react with the ligand 1,2-diaminoethane,

to produce a complex ion.  All the water ligands are replaced and the

coordination number of copper does not change.

Draw the structure of this complex ion, making clear how the ligands form

bonds with the copper ions.

20
(b) (V)
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2 Mark s

When excess concentrated hydrochloric acid is added to 

complex is formed.  Write the equation for this reaction and state the shape of the

complex ion formed. 

Equation:

Shape of complex ion ..................................

20
(b) (VI)
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3 Mark s

The reaction of ammonia with a solution containing ions takes

place in two steps.

 Step 1: A few drops of ammonia are added

 Step 2: An excess of ammonia is added to the products of Step 1.

  Name the types of reaction occurring in the two steps and give the formula 

of the copper complex ion in the final product.

  Step 1 ...........................................................................................................................................

  Step 2 ...........................................................................................................................................

  Final product

Step 1: acid-base / neutralisation

Deprotonation (of complex) / protonation of ammonia

ALLOW (ionic) precipitation (1)

Step 2: Ligand and

Exchange / substitution / replacement

ALLOW

‘Ammonia substitutes for water’

(1)

Final product:

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

ALLOW

[Cu(NH3)4]2+

Round brackets, lack of [ ] brackets

(1)

20
(b) (VII)
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2 Mark s

Describe what you would see in each of the two steps.

 

Step 1: pale blue precipitate/ solid forms

(1)

Step 2: (precipitate dissolves to give) deep / dark blue solution

(1)

Two correct colours with missing states can score (1)

The blue colour in step 2 must be a darker blue than the colour in step one.

e.g. Either pale blue step 1, blue step 2 or blue step 1, dark blue in step 2

20
(b) (VIII)
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3 Mark s

 Explain, in terms of entropy, why the equilibrium constant for the overall

reaction of aqueous copper(II) ions with ammonia is less than the equilibrium

constant when aqueous copper(II) ions react with 1,2-diaminoethane.

M1

No change in number of moles of reactant going to product when ammonia complex forms. There are more moles of product when the diamino complex forms

OR

Increase in the number of moles of product is greater when diamino complex forms

(1)

M2

So greater increase in ΔSsystem / entropy when diamino complex forms

ALLOW

ΔSreaction for ΔSsystem

Reverse argument in M2 based on smaller increase in ΔSsystem when ammonia complex forms

(1)

M3

When ΔSsystem increases,(and ΔSsurrounding remains constant) ΔStotal increases so K increases

Application of:

ΔStotal = ΔSsystem + ΔSsurrounding

ΔStotal = RlnKc

scores M3

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