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Chemistry - Miss Nichols > 23.1 - 23.5 + 24.1 - 24.3 Transition metals > Flashcards

23.1 - 23.5 + 24.1 - 24.3 Transition metals Flashcards

1
Q

Define transition metal

A

A metal that forms at least one stable ion with a partially filled d-shell of electrons

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2
Q

List the general properties of transition metals

A
  • Variable oxidation state
  • Catalysis
  • Complex formation
  • Colour
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3
Q

Define ligand

A

A molecule or ion that has a lone pair of electrons that can be donated to the transition metal

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4
Q

Describe the bonds in transition metal complex ions

A
  • Coordinate bonds
  • Ligands donate e- pairs to the vacant d-orbitals of the transition metal
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5
Q

Define coordination number

A

Number or bonds around the central metal atom or ion

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6
Q

Give the prefixes that denote different numbers of ligands

A
  • 6 ligands = hexa
  • 4 ligands = tetra
  • 2 ligands = di
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7
Q

Define Lewis base

A
  • e- pair donors
  • In complex formations, these are ligands
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8
Q

Define Lewis acid

A
  • e- pair acceptor
  • In complex formations, these are the transition metal ions
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9
Q

Give the parts of naming transition metal complex ions

A
  1. No. of ligands
  2. Type of ligand
  3. Name of the metal
  4. Oxidation number of the metal
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10
Q

Give the naming for different types of ligands

A
  • H20 = aqua
  • Cl- = chloro
  • NH3 = ammine
  • OH- = hydroxo
  • CN- = cyano
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11
Q

Give the naming for negatively charged metal ions

Vanadium
Chromium
Manganese
Iron
Cobalt
Nickel
Copper
Zinc
Silver

A

Vanadium = Vanadate
Chromium = Chromate
Manganese = Manganate
Iron = Ferrate
Cobalt = Cobaltate
Nickel = Nickelate
Copper = Cuprate
Zinc = Zincate
Silver = Argentate

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12
Q

Which ligands form complex ions with a tetrahedral shape?

A
  • Copper
  • Cobalt
  • Most others
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13
Q

Which ligands form complex ions with a square planar shape?

A
  • Platinum
  • Nickel
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14
Q

Define unidente ligands

A
  • Bond through only one donor atom
  • e.g water, ammonia, Cl-
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15
Q

Define bidente ligand

A
  • Bonds through two donor atoms
  • e.g 1,2-diaminoethane, ethandioate ion
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16
Q

Define multidentate ligand

A
  • Bonds through many donor atoms
  • e.g EDTA4-
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17
Q

Describe how complex ions exhibit geometric isomerism

A

Can occur in square planar and octohedral complexes

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18
Q

Describe how complex ions exhibit optical isomerism

A

Can only be exhibited in complexes with bidente or multidente ligands

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19
Q

Describe how transition metal complexes absorb light

A
  • d-orbitals of the transition metal ions split into 2 different energy levels
  • e- absorb energy from visible light
  • Allows e- to move from one d-orbital to another of a higher energy
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20
Q

Explain why transition metal complexes are coloured

A
  • Absorbs certain visble light
  • Complementary colour is reflected
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21
Q

Give the equation to calculate the energy absorbed

A

change in E = hv

E = energy (J)
h = Planck’s constant (J s)
v = frequency of light absorbed (s-1)

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22
Q

Give the factors that effect the amount of energy a transition metal complex absorbs

A
  • The transition metal
  • Type of ligand
  • The co-ordination number
  • Oxidation state of the transition metal
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23
Q

Describe how colorimetry works

A
  • Light source
  • Filter to absorb all irrelevent wave lengths of light
  • Tube with sample
  • Light sensitve cell and meter
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24
Q

What should you do if attempting colourimetry on a sample that is a very pale colour?

A
  • Often hexaaqua ions are very pale
  • Add a different ligand to intensify the colour
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25
Q

When do transition metals exhibit different oxidation states?

A
  • Lower oxidation states exist as simple ions
  • Higher oxidation states exist only when metals are bonded to very electronegative elecments in compound ions
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26
Q

In what conditions is it easier to oxidise a transition metal?

A

Alkaline

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27
Q

In what conditions is it easier to reduce a transition metal?

A

Acidic

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28
Q

Describe why transition metals make good catalysts

A
  • Partially filled d-orbitals
  • Allows them to alternatively accept and reject electrons, cycling between a lower and a higher oxidation state
  • Allows them to help transfer electrons within reactions
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29
Q

Define homogeneous catalyst

A

A catalyst that is in the same phase as the reactants

e.g Breakdown of the ozone by chlorine radicals

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30
Q

Define heterogenous catalyst

A

A catalyst which is in a different phase to the reactants

e.g platinum and rhodium in catalytic converters

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31
Q

Give the equation for the Haber process and its catalyst

A

3H₂ + N₂ ⇌ 2NH₃

  • Fe catalyst
  • Catalyst lasts about 5 years before it becomes poisoned by sulphur and needs replacing
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32
Q

Give the equation for the Contact Process and its catalyst

A

Overall:
2SO₂ + O₂ ⇌ 2SO₃

Steps:
SO₂ + V₂O₅ → SO₃ + V₂O₄
2V₂O₄ + O₂ → 2V₂O₅

  • V₂O₅ catalyst
  • Needed to produce sulphuric acid
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33
Q

Give the equation for making methanol and its catalyst

A

CH₄ + H₂O (steam) → CO + 3H₂
CO + 2H₂ ⇌ CH₃OH

  • Cr₂O₃ catalyst
  • Needed for plastics such as perspex
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34
Q

Describe how adsorption of reactants onto the surface of a heterogenous catalyst increases the rate of reaction

A
  • Brings reactants closer together and increases likelihood of collision
  • May weaken bonds withing the reactant molecules
  • May position the molecule in a favourable orientation for the reaction
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35
Q

Describe how catalytic efficiency is increased

A
  • Increase surface area
  • Spread or impregnate the catalyst onto an inert support medium
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36
Q

Give 2 equations to show how catalytic converters remove NO, CO and unburnt hydrocarbons

A

2CO + 2NO → N₂ + 2CO₂

Hydrocarbon + NO → N₂ + CO₂ + H₂O

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37
Q

Why do heterogenous catalysts need replacing?

A
  • Finely divided catalysts can simply come off the support medium
  • Poisoning - unwanted impurities block the active sites
38
Q

Describe how homogeneous catalysts work

A
  • Can change oxidation state
  • This allows them to form an intermediate species
  • Two reactions with a lower activation energy than the original single reaction
  • Often used for slow reactions which require the collision of two negative ions
39
Q

Describe the oxidation of iodide ions by peroxodisulphate

A

Overall:
2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻

Step by step:
S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

  • Catalysed by Fe
40
Q

Define autocatalysis

A
  • Where one of the products of a reaction is a catalyst for the reaction
  • Catalyst does appear in overall equation
41
Q

Describe the oxidation of ethanedioate ion by manganate (VII) ions

A

2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ + 8H₂O
* Mn²⁺ is catalyst
* MnO₄⁻ = Purple
* Mn²⁺ = Pale pink

Step by step:
* 4Mn²⁺ + MnO₄⁻ + 8H⁺ → 5Mn³⁺ + 4H₂O
* 2Mn³⁺ + C₂O₄²⁻ → 2Mn²⁺ + 2CO₂

42
Q

What colour are copper aqua ions?

43
Q

What colour are iron (II) aqua ions?

44
Q

What colour are iron (III) aqua ions?

45
Q

What colour are aluminium aqua ions?

(Not a transition metal but has similar reactions)

A

Colourless

46
Q

What bonds are present in aqueous metal ions?

A
  • Covalent bond in ligand O-H
  • Coordinate bond between water and metal ion (electron pair donated by O)
47
Q

Describe why metal aqua ions are acidic

A
  • Metal ion pulls electron density away from the H2O ligands
  • Makes them more susceptible to loss of H+
  • Known as hydrolysis
48
Q

Give the equation that explains why metal aqua ions are acidic

A

[M(H₂O)₆]ⁿ⁺ + H₂O ⇌ [M(H₂O)₅OH]ⁿ⁻¹⁺ + H₃O⁺

It is the H₃O⁺ that causes acidity

49
Q

What effects the acidity of metal aqua ions?

A
  • Small, highly charged metal ions are the most acidic
50
Q

Name the 3 main types of reaction that metal-aqua ions undergo

A
  • Hydrolysis (break O-H bond)
  • Ligand substitution (break M-O bond)
  • Redox (Add or remove electrons)
51
Q

Describe hydrolysis reactions

A

Form metal hydroxides

Three possible reagents:
* OH⁻
* dilute NH₃
* CO₃²⁻

52
Q

Describe the products of a hydrolysis reaction with OH⁻

A

M²⁺ = [M(H₂O)₄(OH)₂] + H₂O

M³⁺ = [M(H₂O)₃(OH)₃] + H₂O
Some M³⁺ may react further to give amphoteric hydroxides

53
Q

Describe the products of a hydrolysis reaction with dilute NH₃

A

M²⁺ = [M(H₂O)₄(OH)₂] + NH₄⁺

M³⁺ = [M(H₂O)₃(OH)₃] + NH₄⁺
May get a further reaction if excess NH₃ is used, forming ammine complexes

54
Q

Describe the products of a hydrolysis reaction with CO₃²⁻

A

M²⁺ = MCO₃ (ppt) + H₂O

M³⁺ = [M(H₂O)₃(OH)₃] + CO₂ + H₂O

55
Q

Give the possible reagents for substitution reactions and their products

A

NH₃ (XS or conc) = Ammine complexes

Cl⁻ (conc HCl) = Chloro complexes

56
Q

Give the equation to calculate frequency of light

A

Frequency (Hz) = Speed (m s-1) / Wavelength (nm)

57
Q

Cu²⁺ as aqueous ion

A

[Cu(H₂O)₆]²⁺(aq) + H₂O(l) ⇌ [Cu(H₂O)₅(OH)]⁺(aq) + H₃O⁺(aq)

[Cu(H₂O)₅(OH)]⁺ = Blue solution

58
Q

Cu²⁺, add NaOH(aq) dropwise

A

[Cu(H₂O)₆]²⁺(aq) + 2OH-(aq) ⇌ [Cu(H₂O)₄(OH)₂] (s) + 2H₂O(l)

[Cu(H₂O)₄(OH)₂] = Blue ppt

59
Q

Cu²⁺, add excess NaOH(aq)

A

No reaction

60
Q

Cu²⁺, add NH₃(aq) dropwise

A

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

[Cu(H₂O)₄(OH)₂] = Blue ppt.

61
Q

Cu²⁺, add excess NH₃(aq)

A

[Cu(H₂O)₆]²⁺(aq) + 2NH₃(aq) ⇌ [Cu(NH₃)₄(H₂O)₂] (aq) + 4H₂O(l)

[Cu(NH₃)₄(H₂O)₂] = Very deep blue solution

62
Q

Cu²⁺, add Na₂CO₃ (aq)

A

[Cu(H₂O)₆]²⁺(aq) + CO₃²⁻(aq) → CuCO₃(s) + 6H₂O(l)

CuCO₃ = Green ppt.

63
Q

Cu²⁺, add conc. HCl (aq)

A

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

[CuCl₄]²⁻ = Yellow/green solution

64
Q

Fe²⁺ as aqueous ion

A

[Fe(H₂O)₆]²⁺(aq) + H₂O(l) ⇌ [Fe(H₂O)₅(OH)]⁺(aq) + H₃O⁺(aq)

[Fe(H₂O)₅(OH)]⁺ = Pale green solution

65
Q

Fe²⁺, add NaOH(aq) dropwise

A

[Fe(H₂O)₆]²⁺(aq) + 2OH⁻(aq) ⇌ [Fe(H₂O)₄(OH)₂] (s) + 2H₂O(l)

[Fe(H₂O)₄(OH)₂] = Green/grey ppt.

66
Q

Fe²⁺, add excess NaOH

A

No reaction

67
Q

Fe²⁺, add NH₃ dropwise

A

[Fe(H₂O)₆]²⁺(aq) + 2NH₃(aq) ⇌ [Fe(H₂O)₄(OH)₂] (s) + 2NH₄⁺(aq)

[Fe(H₂O)₄(OH)₂] = Grey/green ppt.

68
Q

Fe²⁺, add excess NH₃

A

No reaction

69
Q

Fe²⁺, add Na₂CO₃

A

[Fe(H₂O)₆]²⁺(aq) + CO₃²⁻(aq) → FeCO₃(s) + 6H₂O(l)

FeCO₃ = Green/grey ppt.

70
Q

Fe²⁺, add conc. HCl

A

No reaction

71
Q

Fe³⁺ as aqueous ion

A

[Fe(H₂O)₆]³⁺(aq) + H₂O(l) ⇌ [Fe(H₂O)₅(OH)]⁺(aq) + H₃O⁺(aq)

[Fe(H₂O)₅(OH)]⁺ = Violet solution

(Can appear somewhat orange due to unreacted [Fe(H₂O)₆]³⁺)

72
Q

Fe³⁺, add NaOH dropwise

A

[Fe(H₂O)₆]³⁺(aq) + 3OH⁻(aq) ⇌ [Fe(H₂O)₃(OH)₃] (s) + 3H₂O(l)

[Fe(H₂O)₃(OH)₃] = Red/brown ppt.

73
Q

Fe³⁺, add excess NaOH

A

No reaction

74
Q

Fe³⁺, add NH₃ dropwise

A

[Fe(H₂O)₆]³⁺(aq) + 3NH₃(aq) ⇌ [Fe(H₂O)₃(OH)₃] (s) + 3NH₄⁺(l)

[Fe(H₂O)₃(OH)₃] = Red/brown ppt.

75
Q

Fe³⁺, add excess NH₃

A

No reaction

76
Q

Fe³⁺, add Na₂CO₃

A

[Fe(H₂O)₆]³⁺(aq) + CO₃²⁻(aq) ⇌ [Fe(H₂O)₃(OH)₃] (s) + 3CO₂(g) + 2H₂O(l)

[Fe(H₂O)₃(OH)₃] = Red/brown ppt.

77
Q

Fe³⁺, add conc. HCl

A

No reaction

78
Q

Al³⁺ as an aqueous ion

A

[Al(H₂O)₆]³⁺(aq) + H₂O(l) ⇌ [Al(H₂O)₅OH]²⁺(aq) + H₃O⁺(l)

[Al(H₂O)₅OH]²⁺ = Colourless solution

79
Q

Al³⁺, add NaOH dropwise

A

[Al(H₂O)₆]³⁺(aq) + 3OH⁻(l) ⇌ [Al(H₂O)₃(OH)₃] (s) + 3H₂O(l)

[Al(H₂O)₃(OH)₃] = White ppt.

80
Q

Al³⁺, add excess NaOH

A

[Al(H₂O)₆]³⁺(aq) + 4OH⁻(l) ⇌ [Al(H₂O)₂(OH)₄]⁻(aq) + 4H₂O(l)

[Al(H₂O)₂(OH)₄]⁻ = Colourless solution

81
Q

Al³⁺, add NH₃ dropwise

A

[Al(H₂O)₆]³⁺(aq) + 3NH₃(aq) ⇌ [Al(H₂O)₃(OH)₃] (s) + 3NH₄⁺(aq)

[Al(H₂O)₃(OH)₃] = White ppt.

82
Q

Al³⁺, add excess NH₃

A

No reaction

83
Q

Al³⁺, add Na₂CO₃

A

[Al(H₂O)₆]³⁺(aq) + CO₃²⁻(aq) ⇌ [Al(H₂O)₃(OH)₃] (s) + 3CO₂(g) + 2H₂O(l)

[Al(H₂O)₃(OH)₃] = White ppt.

84
Q

Al³⁺, add conc. HCl

A

No reaction

85
Q

Explain why an aqueous solution containing [Fe(H2O)6] 3+ ions has a lower pH than an aqueous solution containing [Fe(H2O)6]
2+ ions

A
  • Fe3+ is smaller and has a greater charge
  • Fe3+ ions polarise water molecules more
  • Weakens O-H bond in H2O ligand
  • So more water ligand O-H bonds break, releasing more H+
86
Q

Describe EDTA

A
  • -4 charge
  • Hexadentate
87
Q

What should be given when excess of water is added to ion?

A
  • Hexaaqua ion
  • There is much more of this than of the other ion
88
Q

Give the titration equations that must be memorised

A

MnO₄⁻ + 5e⁻ → Mn²⁺

Cr₂O₇²⁻ + 6e⁻ → Cr³⁺

89
Q

Also learn

A
  • Bidente ligand
  • Cisplatin
90
Q

[Fe(H₂O)₆]³⁺ → [Fe(H₂O)₆]²⁺
What reagent is needed for this reaction?

A

Zn in acid

91
Q

Copper (I) iodide is a white solid. Why?

A
  • Full d-orbital
  • So no d-d transitions
  • Cannot absorb any visible light so appears white

Chemistry - Miss Nichols (17 decks)

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