Brainscape's Knowledge GenomeTM


Top Flashcards (Certified)
All Flashcards

Chemistry aqa > Transition metals > Flashcards

Transition metals Flashcards

(75 cards)

Start Studying
1
Q

Transition Metals

A

Sc - Cu have at least one stable ion and incomplete d sub-level in atoms or ions

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

4 characteristics of transition metals

A
  • complex formation
  • formation of colored ions
  • variable oxidation state
  • catalytic activity
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Why is Zn not a transition metal?

A

Zn 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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

complex

A

is a central metal ion surrounded by ligands

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

ligand

A

An atom, ion or molecule which can donate a lone electron pair

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Co-ordinate bonding

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Co-ordination number

A

The number of co-ordinate bonds formed to a central metal ion

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Ligands can be

A

monodentate (e.g. H2O, NH3, and Cl- ) which can form one coordinate bond per ligand

bidentate (e.g. NH2CH2CH2NH2 and ethanedioate ion C2O4 2- ) which have two atoms with lone pairs and can form two coordinate bonds per ligand

multidentate (e.g. EDTA4- which can form six-coordinate bonds per ligand)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

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

colour

A

deep blue solution

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Reactions with Chloride Ions

A

Addition of a high concentration of chloride ions (from conc HCl or saturated NaCl) to an aqueous ion leads to a ligand substitution reaction.

The Cl ligand is larger than the uncharged H2O and NH3 ligands so therefore ligand exchange can involve a change of co-ordination number

Addition of conc HCl to aqueous ions of Cu and Co leads to a change in coordination number from 6 to 4.

Be careful: If solid copper chloride (or any other metal) is dissolved in water it forms the aqueous [Cu(H2O)6]2+ complex and not the chloride [CuCl4]2- complex.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

[CuCl4]2-

[CoCl4]2-

A

yellow/green solution

blue solution

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Ethane-1-2-diamine

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Ethanedioate

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Copper equations showing the formation of a bidentate ligands

A

[Cu(H2O)6]2+ + 3NH2CH2CH2NH2 —– [Cu(NH2CH2CH2NH2 )3]2+ + 6H2O

[Cu(H2O)6]2+ + 3C2O4 2— [Cu(C2O4 )3]4- + 6H2O

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Partial substitution of ethanedioate ions may occur when a dilute aqueous solution containing ethanedioate ions is added to a solution containing aqueous copper(II) ions.

A

[Cu(H2O)6]2+ + 2C2O4 2—— [Cu(C2O4 )2 (H2O)2]2- + 4H2O

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Equations to show formation of mutidentate complexes

A

[Cu(H2O)6]2+ + EDTA4—— [Cu(EDTA)]2- + 6H2O

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Haem

A

Haem is an iron(II) complex with a multidentate ligand.

Oxygen forms a co-ordinate bond to Fe(II) in hemoglobin, enabling oxygen to be transported in the blood.

CO is toxic to humans because CO can form a strong coordinate bond with hemoglobin. This is a stronger bond than that made with oxygen and so it replaces the oxygen, attaching to the hemoglobin.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

chelate effect

A

The substitution of the monodentate ligand with a bidentate or a multidentate ligand leads to a more stable complex

[Cu(H2O)6]2+ (aq) + EDTA4- (aq)  [Cu (EDTA)]2- (aq) + 6H2O (l)

The copper complex ion has changed from having unidentate ligands to a multidentate ligand.

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.

This chelate effect can be explained in terms of a positive entropy change in these reactions as there are more molecules of products than reactants. Free energy ΔG will be negative as ΔS is positive and ΔH is small. The enthalpy change is small as there are similar numbers of bonds in both complexes

The stability of the EDTA complexes has many applications. 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 to lather

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

A river was polluted with copper(II) ions. 25.0 cm3 sample of the river water was titrated with a 0.0150 mol dm–3 solution of EDTA4– , 6.45 cm3 were required for complete reaction. Calculate the concentration, in mol dm–3 , of copper(II) ions in the river water

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

Shapes of complex ions

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Isomerism in complex ions

cis trans

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Optical isomerism in octahedral complexes

A
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

Color changes arise from changes in

A
  1. oxidation state
  2. co-ordination number
  3. ligand
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

[Co(H2O)6]2+ + 4Cl——[CoCl4]2- + 6H2O

A

pink to blue

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
[Co(NH3 )6]2+ (aq) ----[Co(NH3 )6]3+ (aq) +e +O2
yellow to brown
26
[Co(H2O)6]2+ + 6 NH3 ---- [Co(NH3 )6]2+ + 6H2O
pink to yellow brown
27
How color arises
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 color ΔE = hv v = frequency of light absorbed (unit s-1 or Hz) h= Planck’s constant 6.63 × 10–34 (J s) E = energy difference between split orbitals (J)
28
Changing colour
Changing a ligand or changing the coordination number will alter the energy split between the d- orbitals, changing ΔE and hence change the frequency of light absorbed
29
Compounds without colour
Scandium is a member of the d block. Its ion (Sc3+) hasn't got any d electrons left to move around. So there is not an energy transfer equal to that of visible light In the case of Zn2+ ions and Cu+ ions the d shell is full e.g.3d10 so there is no space for electrons to transfer. Therefore there is not an energy transfer equal to that of visible light
30
Spectrophotometry
If visible light of increasing frequency is passed through a sample of a colored complex ion, some of the light is absorbed. The amount of light absorbed is proportional to the concentration of the absorbing species (and to the distance traveled through the solution). Some complexes have only pale colors and do not absorb light strongly. In these cases, a suitable ligand is added to intensify the color Spectrometers contain a colored filter. The color of the filter is chosen to allow the wavelengths of light through that would be most strongly absorbed by the colored solution Absorption of visible light is used in spectrometry to determine the concentration of colored ions.
31
Spectrophotometry method 5 steps
* Add an appropriate ligand to intensify the colour * Make up solutions of known concentration * Measure absorption or transmission * Plot graph of absorption vs concentration * Measure absorption of unknown and compare
32
Variable oxidation states General trends
* Relative stability of +2 state with respect to +3 state increases across the period * Compounds with high oxidation states tend to be oxidizing agents e.g MnO4 - * Compounds with low oxidation states are often reducing agents e.g V2+ & Fe2+
33
Vanadium
34
[Ag(NH3 )2]+ is used in Tollens’ reagent oxidation and reduction
Red ½ eq: [Ag(NH3 )2]+ + e- ----- Ag +2NH3 Ox ½ eq: CH3CHO + H2O ---- CH3CO2H + 2H+ + 2e
35
The redox titration between Fe2+ with MnO4 – (purple)
MnO4 -(aq) + 8H+ (aq) + 5Fe2+ (aq)----- Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq) purple to colorless
36
Choosing correct acid for manganate titrations
37
A 2.41 g nail made from an alloy containing iron is dissolved in 100 cm3 acid. The solution formed contains Fe(II) ions. 10cm3 portions of this solution are titrated with potassium manganate (VII) solution of 0.02M. 9.80cm3 of KMnO4 were needed to react with the solution containing the iron. Calculate the percentage of iron by mass in the nail
MnO4 - (aq) + 8H+ (aq) + 5Fe2+ ---- Mn2+ (aq) + 4H2O + 5Fe3+
38
manganate titrations with hydrogen peroxide
Ox H2O2---- O2 + 2H+ + 2e- Red MnO4 -(aq) + 8H+ (aq) + 5e- --- Mn2+ (aq) + 4H2O Overall 2MnO4 -(aq) + 6H+ (aq) + 5H2O2 ---- 5O2 + 2Mn2+ (aq) + 8H2O
39
Manganate titration with ethanedioate
Ox C2O4 2- --- 2CO2 + 2e- Red MnO4 -(aq) + 8H+ (aq) + 5e- --- Mn2+ (aq) + 4H2O Overall 2MnO4 -(aq) + 16H+ (aq) + 5C2O4 2-(aq)--- 10CO2 (g) + 2Mn2+(aq) + 8H2O(l)
40
iron (II) ethanedioate formation equation
MnO4 -(aq) + 8H+ (aq) + 5Fe2+ ----- Mn2+ (aq) + 4H2O + 5Fe3+ 2MnO4 -(aq) + 16H+ (aq) + 5C2O4 2- ---- 10CO2 + 2Mn2+ (aq) + 8H2O So overall 3MnO4 -(aq) + 24H+ (aq) + 5FeC2O4 --- 10CO2 + 3Mn2+ (aq) + 5Fe3+ + 12H2O
41
A 1.412 g sample of impure FeC2O4 .2H2O was dissolved in an excess of dilute sulfuric acid and made up to 250 cm3 of solution. 25.0 cm3 of this solution decolourised 23.45 cm3 of a 0.0189 mol dm–3 solution of potassium manganate(VII). Calculate the percentage by mass of FeC2O4 .2H2O in the original sample
42
Catalysis
Catalysts increase reaction rates without getting used up. They do this by providing an alternative route with a lower activation energy
43
heterogeneous catalyst
A heterogeneous catalyst is in a different phase from the reactants
44
homogeneous catalyst
A homogeneous catalyst is in the same phase as the reactants
45
Mechanism of heterogeneous catalyst
Adsorption of reactants at active sites on the surface may lead to catalytic action. The active site is the place where the reactants adsorb on to the surface of the catalyst. This can result in the bonds within the reactant molecules becoming weaker, or the molecules being held in a more reactive configuration. There will also be a higher concentration of reactants at the solid surface so leading to a higher collision frequency
46
Strength of adsorption
The strength of adsorption helps to determine the effectiveness of the catalytic activity. Some metals e.g. W have too strong adsorption and so the products cannot be released. Some metals e.g. Ag have too weak adsorption, and the reactants do not adsorb in high enough concentration. Ni and Pt have about the right strength and are most useful as catalysts.
47
Steps in Heterogeneous Catalysis 4 steps
1. Reactants form bonds with atoms at active sites on the surface of the catalyst (adsorbed onto the surface) 2. As a result bonds in the reactants are weakened and break 3. New bonds form between the reactants held close together on catalyst surface. 4. This in turn weakens bonds between product and catalyst and product leaves (desorbs).
48
Surface area of a solid catalyst
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 (e.g. Rh on a ceramic support in catalytic converters)
49
Contact Process
V2O5 is used as a catalyst 2SO2 + O2 ---- 2SO3 step 1 SO2 + V2O5 ---- SO3 + V2O4 step 2 2V2O4 + O2 --- 2V2O5
50
manufacture of methanol from carbon monoxide and hydrogen
Cr2O3 catalyst CO + 2H2 ----- CH3OH
51
Haber process
Fe is used as a catalyst in the Haber process N2 + 3H2 ---- 2NH3
52
Poisoning Catalysts
Catalysts can become poisoned by impurities and consequently have reduced efficiency. Poisoning has a cost implication e.g. poisoning by sulfur in the Haber process and by lead in catalytic converters in cars means that catalysts lose their efficiency and may need to be replaced.
53
Homogeneous catalysis
When catalysts and reactants are in the same phase, the reaction proceeds through an intermediate species. The intermediate will often have a different oxidation state to the original transition metal. At the end of the reaction the original oxidation state will reoccur. This illustrates importance of variable oxidation states of transition metals in catalysis
54
Reaction between iodide and persulfate ions
55
Autocatalytic Reaction between Ethanedioate and Manganate ions
56
Constructing a catalyzed mechanism for a reaction The following reaction is catalyzed by Co2+ ions in an acidic solution. SO3 2– + ½ O2 ---- SO4 2– Write a mechanism for the catalyzed reaction by writing two equations involving Co 2+ and Co3+
57
Silver Chemistry
Silver behaves like the transition metals in that it can form complexes and can show catalytic behavior (although it adsorbs too weakly for many examples). Silver is unlike the transition metals in that it does not form coloured compounds and does not have variable oxidation states. Silver complexes all have a +1 oxidation state with a full 4d subshell (4d10). As it is 4d10 in both its atom and ion, it does not have a partially filled d subshell and nor is a transition metal by definition. It is not therefore able to do electron transitions between d orbitals that enable coloured compounds to occur.
58
Reactions of halides with silver nitrate
Fluorides produce no precipitate Chlorides produce a white precipitate Ag+ (aq) + Cl - (aq) ---- AgCl(s) Bromides produce a cream precipitate Ag+ (aq) + Br- (aq)----AgBr(s) Iodides produce a pale yellow precipitate Ag+ (aq) + I- (aq) ----- AgI(s)
59
Silver chemistry mechanism
60
[Ag(NH3 )2]+ is used in Tollens’ reagent to distinguish between aldehydes and ketones . Aldehydes reduce the silver in the Tollens’ reagent to silver equaitons
Red ½ eq: [Ag(NH3 )2]+ + e- ----- Ag +2NH3 Ox ½ eq: CH3CHO + H2O ------ CH3CO2H + 2H+ + 2e
61
Using silver nitrate to work out formulae of chloride containing complexes
62
Lewis acid
electron pair acceptor
63
Lewis base
electron pair donator
64
(Fe (H2O)6)2+
green
65
Cu(H20)6 2+
blue
66
Al(H2O)6 3+
colorless
67
Fe(H2O)6 3+
violet In solution and Fe(III) appears yellow/brown due to hydrolysis reactions. The violet colour is only really seen in solid hydrated salts that contain these complexes.
68
WHY the acidity of [M(H2O)6]3+ is greater than that of [M(H2O)6]2+
The acidity of [M(H2O)6]3+ is greater than that of [M(H2O)6]2+ in terms of the greater polarising power (charge/size ratio) of the 3+ metal ion. The greater the polarising power, the more strongly it attracts the water molecule. This weakens the O-H bond so it breaks more easily
69
Reaction with limited OH- and limited NH3
70
Reaction with excess OH-
71
Reaction with excess NH3
72
Reactions with Carbonate solution
73
2+ Ion Summary
74
3+ ion summary
75

Chemistry aqa (21 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2025 Bold Learning Solutions. Terms and Conditions