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transition metal Flashcards

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1
Q

why are zinc and scandium not transition metals

A

because they do not produce a stable ion with a half filled D sub shells zinc can only form a +2 ion. In this ion the Zn 2+ has a complete d orbital and so does not meet the criteria of having an incomplete d orbital in one of its compounds.

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. and Zn hasn’t got enough room

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2
Q

Co-ordination number

A

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

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3
Q

monodentate ligands

A

H2O, NH3 and Cl-

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4
Q

bidentate ligands

A

NH2CH2CH2NH2 (Ethane-1-2-diamine) and ethanedioate ion C2O4 2-

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5
Q

multidente ligands

A

EDTA4- which can form six coordinate bonds per ligand

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6
Q

what two monodente ligands are very similar

A

The ligands NH3 and H2O are similar in size and are uncharged. Exchange of the ligands NH3 and H2O occurs without change of co-ordination number

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7
Q

chloride ligand subsitution

A

Addition of conc HCl to aqueous ions of Cu and Co leads to a change in coordination
number from 6 to 4.
[CuCl4]2- yellow/green solution
[CoCl4]2- blue solution
These are tetrahedral in shape.

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8
Q

Partial substitution of ethanedioate

A

ions may occur when a dilute aqueous solution containing ethanedioate ions is added to a solution containing aqueous copper(II) ions. In this reaction four water molecules are replaced and a new complex is formed.

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

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9
Q

what is Haem

A

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

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10
Q

why is carbon monoxide toxic

A

Oxygen forms a co-ordinate bond to Fe(II) in haemoglobin, enabling oxygen to be transported in the blood.
CO is toxic to humans because CO can form a strong coordinate bond with haemoglobin. This is a stronger bond than that made with oxygen and so it replaces the oxygen, attaching to the haemoglobin.

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11
Q

chelate effect

A

The substitution of monodentate ligand with a bidentate or a multidentate ligand leads to a more stable complex. this because of positive entropy change in these reactions as there are more molecules of products than reactants.

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12
Q

stability of the EDTA complex applications

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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13
Q

is the enthalpy change of substitution reactions small

A

yes because the same number of bonds are being made and broken

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14
Q

what forms octahedral shape

A

small ligands (e.g. H2O and NH3).

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15
Q

what forms tetrahedral shape

A

complexes with larger ligands (e.g.Cl- ).

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16
Q

what forms Square planar

A

cisplatin to Cl- ligands and two NH3s

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17
Q

what forms linear complexes

A

Ag+ commonly forms linear complexes e.g. [Ag(NH3)2]+ used as Tollen’s reagent

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18
Q

what show cis-trans

A

octahedral complexes, square planar

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19
Q

what shape shows optical isomerism

A

octahedral complexes

Complexes with 3 bidentate ligands can form two optical
isomers (non-superimposable mirror images).

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20
Q

Colour changes arise from changes in

A
  1. oxidation state,
  2. co-ordination number
  3. ligand.

because 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 hence changes the light that’s transmitted to give substance colour.

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21
Q

How colour arises

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.

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22
Q

VO2 +

A

a yellow solution

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23
Q

VO 2+

A

blue solution

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24
Q

V3+

A

green solution

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25
Q

V2+

A

violet solution

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26
Q

zinc added to vanadium reducing agent

A

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.

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27
Q

tollen’s reagent equation

A

Aldehydes reduce the silver in the Tollens’ reagent to silver.
[Ag(NH3)2]+ + e- —–> Ag +2NH3

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28
Q

Manganate Redox Titration which Fe 2+

A

The redox titration between Fe2+ with MnO4– (purple).

MnO4-(aq) + 8H+ (aq) + 5Fe2+ (aq) —–> Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)

purple to colourless

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29
Q

how can you tell end point of Manganate Redox Titration

A

If the manganate is in the burette then the end point of the titration will be the first permanent pink colour.

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30
Q

acid for manganate ions

A

needs 8H+

Only use dilute sulfuric acid for manganate titrations.

Insufficient volumes of sulfuric acid will mean the solution is not acidic enough and MnO2 will be produced instead of Mn2+. Using a weak acid like ethanoic acid would have the same effect

MnO4 -(aq) + 4H+(aq) + 3e- —-> MnO2 (s) + 2H2O

The brown MnO2 will mask the colour change and lead to a greater (inaccurate) volume of manganate being used in the titration.

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31
Q

why can’t HCl be the acid

A

It cannot be conc HCl as the Cl- ions would be oxidised to Cl2 by MnO4-

because
Eo MnO4-/Mn2+ > Eo Cl2/Cl-

This would lead to a greater volume of manganate being used and poisonous Cl2 being produced.

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32
Q

Why can’t you use nitric acid

A

It oxidises Fe2+ to Fe3+

This would lead to a smaller volume of manganate being used.

33
Q

manganate and hydrogen peroxide titration

A

the hydrogen peroxide is oxidised

manganate reduced

34
Q

manganate and ethanedioate

A

ethanedioate oxidised

manganate reduced

35
Q

how do heterogeneous catalysts work

A
  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).
36
Q

strength of absorption

A

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.

37
Q

best catalysts

A

Ni and Pt have about the right strength and are most useful as catalysts.

38
Q

Examples of heterogeneous catalysts

A

V2O5 is used as a catalyst in the Contact process.

Fe is used as a catalyst in the Haber process

Cr2O3 catalyst is used in the manufacture of methanol from carbon monoxide and hydrogen.

39
Q

Contact process equation

A

Overall equation :2SO2 + O2 –>2SO3

step1: SO2 +V2O5 —>SO3 + V2O4
step 2: 2V2O4 + O2 —> 2V2O5

40
Q

haber process equation

hat
ber process equation

A

N2 + 3H2 —-> 2NH3

41
Q

manufacture of methanol from carbon monoxide and hydrogen equation

A

CO + 2H2 —–>CH3OH

42
Q

Poisoning catalysts

A

Catalysts can become poisoned by impurities and consequently have reduced efficiency. Poisoning has a cost implication

43
Q

leaded petrol and catalytic converter

A

Leaded petrol cannot be used in cars fitted with a catalytic converter since lead strongly adsorbs onto the surface of the catalyst.

44
Q

importance of variable oxidation states for homogenous catalysts

A

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.

45
Q

Sc ion

A

Sc 3+

46
Q

Ti ion

A

Ti 3+

47
Q

V ion

A

V 3+

48
Q

Cr ion

A

Cr 3+

49
Q

Mn ion

A

Mn 2+

50
Q

Fe ion

A

can form Fe2+ (because it readily oxidises) and Fe3+

51
Q

Co ion

A

Co 2+

52
Q

Ni ion

A

Ni 2+

53
Q

Cu ion

A

Cu 2+

54
Q

Zn ion

A

Zn 2+

55
Q

ethandioate structure

A
56
Q

EDTA titrations

A

EDTA always in 1:1 ratio with metal

57
Q

EDTA charge

A

4-

58
Q

manganate ion used in titration

A

MnO4 -

59
Q

oxidation of Fe 2+ with manganate ions equation

A

MnO4-(aq) + 8H+ (aq) + 5Fe2+ (aq) ——–> Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq

60
Q

MnO4 - reduced equation

A

MnO4 - +8H+ + 5e- —–> Mn2+ + 4H20

61
Q

why is the reaction between manganate and ethandioate slow

A

because its two negative ions reacting with each other

62
Q

2 examples of homogenous catalyst

A

Reaction between iodide and persulfate ions

Autocatalytic reaction between ethanedioate and manganate ions

63
Q

Reaction between iodide and persulfate ions

A

The reaction between I- and S2O8 2- catalysed by Fe2+

The uncatalysed reaction is very slow because the reaction needs a collision between two negative ions. Repulsion between the ions is going to hinder this – meaning high activation energy.

64
Q

where does something electrode potential have to lie if it wants to acts as a homogenous catalyst

A

it needs to be in-between the two reactant as it needs to be both oxidised and reduced

65
Q

original iodide ion and presulfate reaction

A

S2O8 2- + 2I- —-> 2SO4 2- + I2

66
Q

catalysed alternative route iodide ion and presulfate reaction

A

stage 1: S2O8 2- + 2Fe 2+ —-> 2SO4 2- + 2Fe 3+

stage 2: 2I- + 2Fe 3+ ——> I2 + 2Fe 2+

(these steps are interchangeable do Fe3+ and Fe2+ can act as catalysts)

67
Q

Autocatalytic reaction between ethanedioate and manganate ions

A

The autocatalysis by Mn2+ in titrations of C2O4 2- with MnO4-

(the product is also the catalyst)

68
Q

ethanedioate and manganate ions catalysed route

A

Step 1: 4Mn2+ + MnO4- + 8 H+ ——> 5Mn3+ + 4 H2O

Step 2: 2Mn3+ + C2O42- ——> 2Mn2+ + 2 CO2

69
Q

how do follow the reaction rate of this reaction

A

This can be done by removing samples at set times and titrating to work out the concentration of MnO4- .
It could also be done by use of a spectrometer measuring the intensity of the purple colour.

This method has the advantage that it does not disrupt the reaction mixture, using up the reactants and it leads to a much quicker determination of concentration.

70
Q

construction a catalyst mechanism for a reaction

A
  1. split overall equation into two half equation
  2. write equations of catalyst being reduced and oxidised
  3. combines oxidised catalyst with reduced half equation and rise versa
71
Q

how does silver act as a transition metal

A

Silver behaves like the transition metals in that it can form complexes and can show catalytic behaviour (although it adsorbs too weakly for many examples).

72
Q

how does silver not act as a transition metal

A

Silver is unlike the transition metals in that it does not form coloured compounds and does not have variable oxidation states. It always forms 1+ complex. its subshell is 4d10 in both its atom and ion, it does not have a partially filled d subshell and so is not a transition metal by definition. It is not therefore able to do electron transitions between d orbitals that enable coloured compounds to occur.

73
Q

Silver chloride dissolves in dilute ammonia to form a complex ion equation

A

AgCl(s) + 2NH3(aq) ——> [Ag(NH3)2]+ (aq) + Cl- (aq)
Colourless solution

74
Q

Silver bromide dissolves in dilute ammonia to form a complex ion equation

A

AgBr(s) + 2NH3(aq) —–> [Ag(NH3)2]+ (aq) + Br - (aq)

75
Q

Co(NH3)6Cl3 react with silver chloride

A

3+ because all the chlorides are on the outsides

76
Q

Cr(NH3)5Cl3 react with silver chloride

A

2+ because 2 chlorides are on the outside

77
Q

Cr(NH3)4Cl react with

A

1+ because one chloride is on the outside

78
Q
A

Chemistry (10 decks)

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