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AQA A-Level Inorganic Chemistry > [3.2.5] Transition Metals > Flashcards

[3.2.5] Transition Metals Flashcards

General properties of transition metals, Substitution reactions, Shapes of complex ions, Formation of coloured ions, Variable oxidation states & Catalysts.

1
Q

What do transition metal characteristics of elements Sc -> Cu arise from?

What do the characteristic properties include?

A
  • Transition metal characteristics of elements Sc -> Cu arise from an incomplete d sub-level in atoms or ions.
  • These characteristics include:
    • Complex formation.
    • Formation of coloured ions.
    • Variable oxidation state.
    • Catalytic activity.
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2
Q

Write the electronic configuration for elements Sc -> Zn.

If these atoms were to form ions, which subshell do they lose electrons from first?

A
  • When forming ions, lose 4s before 3d.
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3
Q

Why is zinc not a transition metal?

A
  • Zinc can only form a +2 ion.
  • In this ion, the Zn²⁺ has a complete d orbital and so does not meet the criteria of having an incomplete d orbital in one of its compounds.
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4
Q

What is a ligand?

A
  • An atom, ion or molecule which can donate a lone electron pair.
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5
Q

What is a complex?

A
  • A central metal ion surrounded by ligands.
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6
Q

What is coordinating bonding?

A
  • Co-ordinate bonding is when the shared pair of electrons in the covalent bond come from only one of the bonding atoms.
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7
Q

What is a coordination number?

A
  • The number of coordinate bonds formed to a central metal ion
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8
Q

What is a monodentate ligand? Give some examples.

A
  • Ligands that have 1 atom with a lone pair only form one coordination bond per ligand.
  • This includes: H₂O, NH₃ & Cl⁻.
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9
Q

What is a bidentate ligand? Give some examples.

A
  • Ligands that have two atoms with lone pairs and can form two coordinate bonds per ligand.
  • This includes: NH₂CH₂CH₂NH₂ and C₂O₄²⁻.
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10
Q

What is a multidentate ligand? Give an example.

A
  • Ligands that have 3+ atoms with lone pairs that can form 3+ coordinate bonds per ligand.
  • This includes: EDTA⁴⁻ (which can form six coordinate bonds per ligand).
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11
Q

Exchange of the ligands NH₃ & H₂O occurs without the change of coordination number. Write an equation to demonstrate this using Co²⁺ & Cu²⁺ complexes.

A
  • [Co(H₂O)₆]²⁺ (aq) + 6NH₃ (aq) -> [Co(NH₃)₆]²⁺ (aq) + 6H₂O (l)
  • [Cu(H₂O)₆]²⁺ (aq) + 4NH₃ (aq) -> [Cu(NH₃)₄(H₂O)₂]²⁺ (aq) + 4H₂O (l)
    • The substitution is incomplete with copper.
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12
Q

Why can the exchange of the ligand H₂O by Cl⁻ involve a change of coordination number? Write an equation to demonstrate this using Co²⁺, Cu²⁺ & Fe³⁺ complexes.

A
  • The Cl⁻ ligand is larger than the uncharged H₂O & NH₃ ligands so therefore ligand exchange can involve a change of coordination number.
  • [Cu(H₂O)₆]²⁺ + 4Cl⁻ -> [CuCl₄]²⁻ + 6H₂O
  • [Co(H₂O)₆]²⁺ + 4Cl⁻ -> [CoCl₄]²⁻ + 6H₂O
  • [Fe(H₂O)₆]³⁺ + 4Cl⁻ -> [FeCl₄]⁻ + 6H₂O
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13
Q

Write an equation to demonstrate the exchange of monodentate ligand H₂O by bidentate ligand ethane-1-2-diamine (NH₂CH₂CH₂NH₂) using Cu²⁺ as an example.

A
  • [Cu(H₂O)₆]²⁺ + 3NH₂CH₂CH₂NH₂ -> [Cu(NH₂CH₂CH₂NH₂)₃]²⁺ + 6H₂O.
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14
Q

Write an equation to demonstrate the exchange of monodentate ligand H₂O by bidentate ligand ethanedioate (C₂O₄²⁻) using a Cu²⁺ complex.

A
  • [Cu(H₂O)₆]²⁺ + 3C₂O₄²⁻ -> [Cu(C₂O₄)₃]⁴⁻ + 6H₂O.
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15
Q

Write an equation to demonstrate the exchange of monodentate ligand H₂O by multidentate ligand EDTA⁴⁻ using a Cu²⁺ complex.

A
  • [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ -> [Cu(EDTA)]²⁻ + 6H₂O.
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16
Q

What can we use EDTA for?

A
  • It can be added to rivers to remove poisonous heavy metal ions as the EDTA complexes are not toxic.
  • It is in many shampoos to remove calcium ions present in hard water, so helping lathering.
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17
Q

What is haem?

How does it enable oxygen to be transported in the blood?

Why is CO toxic to humans?

A
  • Haem is an iron (II) complex with a multidentate ligand.
  • Oxygen forms a coordinate bond to Fe (II) in haemoglobin, enabling oxygen to be transported in the blood.
  • CO can form a strong coordinate bond with haemoglobin which is stronger than that made with oxygen so it replaces the oxygen.
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18
Q

What is the effect that describes when bidentate and multidentate ligands replace monodentate ligands from complexes to make a more stable complex?

What can it be explained in terms of?

A
  • Chelate effect.
  • It can be explained in terms of a positive entropy change as it results in there being more molecules of products than reactants.
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19
Q

What can we say about entropy in this reaction: [Cu(H₂O)₆]²⁺ + EDTA⁴⁻ -> [Cu(EDTA)]²⁻ + 6H₂O?

A
  • In this reaction, there is an increase in entropy because there are more moles of products than reactants (from 2 to 7), creating more disorder.
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20
Q

What can we say about entropy in this reaction: [Cu(H₂O)₆]²⁺ + 3NH₂CH₂CH₂NH₂ -> [Cu(NH₂CH₂CH₂NH₂)₃]²⁺ + 6H₂O?

A
  • This reaction has an increase in entropy because of the increase in moles from 4 to 7 in the reaction.
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21
Q

What shape of molecule do transition metals commonly form with small ligands such as H₂O & NH₃?

Draw the shape of the complex ion [Ni(NH₃)₆]²⁺.

A
  • Octahedral.
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22
Q

What shape of molecule do transition metals commonly form with larger ligands such as Cl⁻?

Draw the shape of the complex ion [CuCl₄]²⁻.

A
  • Tetrahedral.
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23
Q

Draw the shape of the complex [Cr(NH₂CH₂CH₂NH₂)₃]³⁺. What shape is this complex?

A
  • Octahedral.
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24
Q

Draw the shape of the complex [Cr(C₂O₄)₃]³⁻. What shape is this complex?

A
  • Octahedral.
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25
Q

Octahedral complexes can display cis-trans isomerism with monodentate ligands. Demonstrate this isomerism with the complex [Cr(H₂O)₄Cl₂]⁺.

A
26
Q

Octahedral complexes can display optical isomerism with bidentate ligands. Demonstrate this isomerism with the complex [Ni(NH₂CH₂CH₂NH₂)₃]²⁺.

A
27
Q

What shape is cisplatin? Draw the molecule.

A
  • Square planar.
28
Q

Square planar complexes can display cis-trans isomerism. Demonstrate this with the complex Ni(NH₃)₂Cl₂.

A
29
Q

What complex does Ag⁺ commonly form? Draw the shape of the complex [Ag(NH₃)₂]+.

A
  • Linear.
30
Q

How does colour arise?

A
  • Colour arises from electronic transitions from the ground state to excited states between different d orbitals.
  • A portion of visible light is absorbed to promote d electrons to higher energy levels.
  • The light that is not absorbed is transmitted to give the substance colour.
31
Q

What formula represents the energy difference between the ground state and the excited state of the d electrons?

A

ΔE = hv = hc/λ

ΔE = energy difference between split orbitals (J)
h = Planck’s constant (6.64 x 10⁻³⁴) (J s)
v = frequency of light absorbed (s⁻¹ or Hz)
c = speed of light (3.00 x 10⁸) (m s⁻¹)
λ = wavelength of light absorbed (m)

32
Q

What do colour changes arise from in transition metals?

Why does this happen?

A
  1. Changes in oxidation state.
    • e.g. [Co(NH₃)₆]²⁺ (YELLOW) -> [Co(NH₃)₆]³⁺ (BROWN) + e⁻
  2. Changes in coordination number.
    • e.g. [Co(H₂O)₆]²⁺ (PINK) + 4Cl⁻ -> [CoCl₄]²⁻ (BLUE) + 6H₂O
  3. Changes in ligand.
    • e.g. [Co(H₂O)₆]²⁺ (PINK) + 6NH₃ -> [Co(NH₃)₆]²⁺ (YELLOW BROWN) + 6H₂O
  • Changing a ligand, coordination number or oxidation state will alter the energy split between the d-orbitals, thus changing ΔE and hence the frequency of light absorbed.
33
Q

Why don’t the ions Sc³⁺, Zn²⁺ & Cu⁺ have any colour?

A
  • Sc³⁺ hasn’t got any d electrons left to move around so there is not an energy transfer equal to that of visible light.
  • Zn²⁺ and Cu⁺ have full d shells so there is no space for electrons to transfer therefore, there is not an energy transfer equal to that of visible light.
34
Q

How is the absorption of visible light used in spectroscopy?

A
  • If visible light of increasing frequency is passed through a sample of a coloured complex ion, some of the light is absorbed.
  • The amount of light absorbed is proportional to the concentration of the absorbing species.
  • Some complexes have only pale colours and do not absorb light strongly -> a suitable ligand is added to intensify the colour.
35
Q

How can a colourimeter be used to determine the concentration of coloured ions in solution?

A
  1. Make up solutions of known concentration.
  2. Measure absorption or transmission.
  3. Plot a graph of absorption vs concentration.
  4. Measure absorption of unknown and compare to graph.
36
Q

What is the redox potential for a transition metal ion changing from a higher to a lower oxidation state influenced by?

A
  • pH and by the ligand.
37
Q

What is the result of adding zinc to vanadium (V) in acidic solution?

A
  • The addition of zinc to the vanadium (V) in acidic solution will reduce the vanadium down through each successive oxidation state and the colour will successively change from yellow to blue to green to violet.
    • VO₂⁺ = +5 oxidation state = yellow solution.
    • VO²⁺ = +4 oxidation state = blue solution.
    • V³⁺ = +3 oxidation state = green solution.
    • V²⁺ = +2 oxidation state = violet solution.
38
Q

What is used to distinguish between aldehydes and ketones?

A
  • The reduction of [Ag(NH₃)₂]⁺ to metallic silver.
39
Q

What is the half-equation that represents the reaction between Fe²⁺ and MnO₄⁻?

A
40
Q

In the redox titration between Fe²⁺ and MnO₄⁻, what is the colour change that occurs?

A
41
Q

What acid should be used in manganate titrations? Why is it important to ensure that enough of this acid is used?

A
42
Q

Using the half-equations below and their E° values, explain why HCl cannot be used in manganate titrations.

A
43
Q

Using the half-equations below and their E° values, explain why nitric acid cannot be used in manganate titrations.

A
44
Q
A
45
Q

What is the half-equation that represents the reaction between hydrogen peroxide and MnO₄⁻?

A
46
Q

What is the half-equation that represents the reaction between ethanedioate and MnO₄⁻?

A
47
Q

What is the half-equation that represents the reaction between iron (II) ethanedioate and MnO₄⁻?

A
48
Q
A
49
Q

What do catalysts do?

A
  • Catalysts increase reaction rates without getting used up.
  • They do this by providing an alternative route with a lower activation energy.
50
Q

What is a heterogenous catalyst?

A
  • Heterogenous catalysts are in a different phase from the reactants.
  • They are usually solids whereas, the reactants are gaseous or in solution.
51
Q

Describe the steps in heterogeneous catalysis.

A
  1. Adsorption of reactants at active sites on the surface of the catalyst.
  2. Results in the bonds within the reactant molecule weakening and thus breaking.
  3. New bonds form between the reactants held close together on catalyst surface.
  4. This weakens the bonds between product and catalyst and the product desorbs.
52
Q

How does the strength of adsorption help to determine the effectiveness of the catalytic activity?

A
  • Some metals have too strong adsorption and so the products cannot be released.
  • Some metals have too weak adsorption and the reactants do not adsorb in high enough concentration.
53
Q

How can we improve the effectiveness of a catalyst?

A
  • Increasing the surface area of a solid catalyst will improve its effectiveness.
    • A support medium is often used to maximise the surface area and minimise the cost.
54
Q

Explain, with the aid of equations, how V₂O₅ is used as a heterogeneous catalyst in the Contact process.

A

STEP 1

SO₂ + V₂O₅ -> SO₃ + V₂O₄

STEP 2

2V₂O₄ + O₂ -> 2V₂O₅

OVERALL

2SO₂ + O₂ -> 2SO₃

55
Q

What heterogeneous catalyst is used in the Haber Process?

A
  • Fe.
56
Q

What can negatively impact the efficiency of heterogeneous catalysts? How does this impact cost?

A
  • Catalysts can become poisoned by impurities and thus have reduced efficiency.
  • This has a cost implication as catalysts would have to be replaced.
57
Q

What catalyst is used in the manufacture of methanol from carbon monoxide and hydrogen?

A
  • Cr₂O₃.
58
Q

What is a homogenous catalyst?

A
  • When catalysts and reactants are in the same phase so the reaction proceeds through an intermediate species.
59
Q

What is the importance of variable oxidation states in homogenous catalysis?

A
  • The intermediate will often have a different oxidation state than the original transition metal.
  • At the end of the reaction, the original oxidation state will reoccur.
  • This illustrates the importance of variable oxidation states of transition metals in catalysis.
60
Q

Explain, with the aid of equations, how Fe²⁺ ions catalyse the reaction between I⁻ and S₂O₈²⁻.

Why is the uncatalysed reaction very slow?

A

STEP 1

S₂O₈²⁻ + 2Fe²⁺ -> 2SO₄²⁻ + 2Fe³⁺

STEP 2

2I⁻ + 2Fe³⁺ -> 2Fe²⁺ + I₂

OVERALL

S₂O₈²⁻ + 2I⁻ -> 2SO₄²⁻ + I₂

WHY IS THE UNCATALYSED REACTION SLOW?

  • The reaction needs a collision between two negative ions.
  • Repulsion between the ions is going to hinder this resulting in a high activation energy.
  • In the catalysed reaction, the individual stages involve the collision between positive and negative ions and so have lower activation energies.
61
Q

Explain, with the aid of equations, how Mn²⁺ ions autocatalyse the reaction between C₂O₄²⁻ and MnO₄⁻.

Why is the initial uncatalysed reaction slow?

How do the Mn²⁺ ions act as an autocatalyst?

A

STEP 1

4Mn²⁺ + MnO₄⁻ + 8H⁺ -> 5Mn³⁺ + 4H₂O

STEP 2

2Mn³⁺ + C₂O₄²⁻ -> 2Mn²⁺ + 2CO₂

OVERALL

2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ -> 2Mn²⁺ + 10CO₂ + 8H₂O

WHY IS THE UNCATALYSED REACTION SLOW?

  • It’s slow because the reaction is a collision between two negative ions which repel each other leading to high activation energy.

HOW DO THE Mn²⁺ IONS ACT AS AN AUTOCATALYST?

  • The Mn²⁺ ions produced act as an autocatalyst and so the reaction speeds up because they bring about the alternative reaction route with a lower activation energy.
  • The reaction slows as the MnO₄⁻ concentration drops.

AQA A-Level Inorganic Chemistry (6 decks)

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