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AQA A level Chemistry HJ > Transition Metals > Flashcards

Transition Metals Flashcards

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1
Q

why do the elements from Sc to Cu have the characteristics of transition metals?

A

they have an incomplete d sub-level in atoms or ions

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2
Q

why do the elements from Sc to Cu have the characteristics of transition metals?

A

they have an incomplete d sub-level in atoms or ions

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3
Q

what are the characteristics of transition metals?

A
  • complex formation
  • formation of coloured ions
  • variable oxidation states
  • catalytic activity
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4
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

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5
Q

define complex

A

is a central metal ion surrounded by ligands

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6
Q

define ligand

A

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

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7
Q

what is co-ordinate bonding?

A

when the shared pair of electrons in the covalent bond come from only one of the bonding atoms

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8
Q

how is coordinate bonding involved in transition metals?

A

when the shared pair of electrons in the covalent bond come from only one of the bonding atoms

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9
Q

define co- ordinate number

A

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

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10
Q

what are the 3 types of ligands?

A
  • monodentate
  • bidentate
  • multidentate
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11
Q

what is meant by monodentate + give some example?

A

they can form one coordinate bond per ligand
e.g.=
H2O, NH3 and Cl

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12
Q

what is meant by bidentate + give some example?

A

they have two atoms with lone pairs and can form two coordinate bonds per ligand
e.g=
NH2CH2CH2NH2 and ethanedioate ion C2O4
2-

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13
Q

what is meant by multidentate + give some example?

A

which can form six coordinate bonds per ligand
e.g=
EDTA 4-

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14
Q

describe what what happens with the exchange of the ligands NH3 and H2O in the reaction with [Co(H20)6] 2+ and [Cu(H2O)6] 2+
& write the equations

A

-there is no change in the co-ordination number
[Co(H2O)6]2+ (aq) + 6NH3 (aq) -> [ Co(NH3)6]2+ (aq) + 6H2O (l)

-and with the substitution is also incomplete
[Cu(H2O)6]2+ (aq) + 4NH3 (aq) -> [ Cu(NH3)4 )H2O)2]2+ (aq) + 4H2O (l)

-Cu becomes [ Cu(NH3)4 )H2O)2]2+ deep blue solution

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15
Q

describe what what happens with the exchange of the ligands NH3 and H2O in the reaction between [Co(H20)6] 2+
& write the equations

A

-there is no change in the co-ordination number (e.g. CO2+ and Cu2+

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16
Q

what are the characteristics of transition metals?

A
  • complex formation
  • formation f coloured ions
  • variable oxidation states
  • catalytic activity
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17
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

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18
Q

define complex

A

is a central metal ion surrounded by ligands

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19
Q

define ligand

A

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

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20
Q

what is co-ordinate bonding?

A

when the shared pair of electrons in the covalent bond come from only one of the bonding atoms

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21
Q

how is coordinate bonding involved in transition metals?

A

when the shared pair of electrons in the covalent bond come from only one of the bonding atoms

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22
Q

define co- ordinate number

A

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

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23
Q

what are the 3 types of ligands?

A
  • monodentate
  • bidentate
  • multidentate
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24
Q

what is meant by monodentate + give some example?

A

they can form one coordinate bond per ligand
e.g.=
H2O, NH3 and Cl

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25
Q

what is meant by bidentate + give some example?

A

they have two atoms with lone pairs and can form two coordinate bonds per ligand
e.g=
NH2CH2CH2NH2 and ethanedioate ion C2O4
2-

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26
Q

what is meant by multidentate + give some example?

A

which can form six coordinate bonds per ligand
e.g=
EDTA 4-

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27
Q

what are some similarities between NH3 and H2O and what types of ligands are they?

A
  • they are similar in size and uncharged

- they are monodentate ligands

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28
Q

describe what what happens with the exchange of the ligands NH3 and H2O in the reaction between [Co(H20)6] 2+
& write the equations

A

-there is no change in the co-ordination number (e.g. CO2+ and Cu2+

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29
Q

what happens when you add a high conc of chloride ions to an aqueous ion?

A

it leads to a ligand substitution reaction

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30
Q

where does the the high conc of chloride ions needed for ligand substitution reactions come from?

A

from conc HCl or saturated NaCl

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31
Q

what happens when you add conc HCl to aqueous ions of Cu and Co?
+ colours and shape

A

-it leads to a change in coordination number from 6 to 4
[CuCl4] 2- is a yellow/ green solution
[CoCl4] is a blue solution
-these are tetrahedral in shape

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32
Q

why does the substitution of Cl- ligands lead to a change in coordination number but the substitution of H20 and NH3 not?

A

the Cl- ligand is larger than the uncharged H2O and NH3 ligands

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33
Q

write the equation for the addition of a high conc of Cl- ions to [Cu (H2O)6]2+

A

[Cu(H2O)6] 2+ + 4Cl- -> [CuCl4]2- + 6H2O

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34
Q

write the equation for the addition of a high conc of Cl- ions to [Co (H2O)6]2+

A

[Co(H2O)6] 2+ + 4Cl- -> [CoCl4]2- + 6H2O

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35
Q

write the equation for the addition of a high conc of Cl- ions to [Fe (H2O)6]2+

A

[Fe(H2O)6] 3+ + 4Cl- -> [FeCl4]- + 6H2O

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36
Q

what happens if solid copper chloride (or any other metal is dissolved in water?

A

it forms the aqeuous [Cu (H2O)6] 2+ complex and not the chloride [CuCl4]2- complex

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37
Q

what are some examples of common bidentate ligands

A

Ethane-1-2-diamine

Ethanedioate

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38
Q

draw the complex with the bidentate ligand: Ethane-1-2-diamine

  • with the formula [Cr(NH2CH2CH2NH2)3] 3+
  • how many bidentate ligands does it have and how many times do they bond to the metal ion?
  • what is its coordination number?
  • write the equation for this reaction
A

-There are 3 bidentate ligands in this complex
each bonding in twice to the metal ion
-It has a coordination number of 6

[Cu(H2O)6]2+ + 3NH2CH2CH2NH2 -> [Cu (NH2CH2NH2)3] 2+ + 6H2O

page 2
https://chemrevise.files.wordpress.com/2022/01/2.5-revision-guide-transition-metals-aqa.pdf

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39
Q

draw the complex with the bidentate ligand: Ethanedioate

  • with the formula [Cr(C2O4)3]3-
  • what is the bond angle and shape?
A

-octahedral shape with bond angles of 90 degrees

[Cu(H2O)6]2+ + 3C2O42- -> [Cu(C2O4)3]4- + 6H2O

page 2
https://chemrevise.files.wordpress.com/2022/01/2.5-revision-guide-transition-metals-aqa.pdf

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40
Q

write the short hand for the formation of ions for the elements between Sc and Cu (including these)

A 
Sc 3+ [Ar] 4s03d0 
Ti 3+ [Ar] 4s03d1 
V3+ [Ar] 4s03d2
Cr 3+ [Ar] 4s03d3 
Mn 2+ [Ar] 4s03d5
Fe 3+ [Ar] 4s03d5 
Co 2+ [Ar] 4s03d7
Ni 2+ [Ar] 4s03d8
Cu 2+ [Ar] 4s03d9 
Zn 2+ [Ar] 4s03d10
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41
Q

what happens when you add dilute aqeuous solution containing ethandioate ions to a solution containing aqeous copper (II) ions?

-include the equation

A
  • partial substitution of ethanedioate ions may occur
  • in this reaction four water molecules are replaced and a new complex is formed

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

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42
Q

describe the structure and function haem

A

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

  • 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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43
Q

draw the displayed formula for EDTA 4- anion

A

page 2

https://chemrevise.files.wordpress.com/2022/01/2.5-revision-guide-transition-metals-aqa.pdf

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44
Q

write the equation showing the formation of multidentate complexes using [Cu(H2O)6]2+ + EDTA

A

[Cu(H2O)6]2+ + EDTA -> [Cu(EDTA )]2- +6H2O

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45
Q

what is meant by the chelate effect?

A
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46
Q

what are the 2 types of stereoisomerism?

A

cis - trans isomerism and optical isomerism

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47
Q

what are some common complex ions and their bond angles?

A
  • octahedral - 90 degrees
  • tetrahedral - 109.5 degrees
  • square planar - 90 degrees
  • linear- 180 degrees
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48
Q

what type of isomerism is shown by octahedral complexes with bidentate ligands?

A

optical isomerism - the 2 isomers are mirror images of each other

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49
Q

what is the shape of complexes which form larger ligands such as Cl-?

A

when complexes form with larger ligands such as Cl-, they form tetrahedral complexes with a bond angle of 109.5 degrees.

-tetrahedral complexes can also show optical isomerism

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50
Q

what is the shape of the complexes formed with the ligands Platinum and Nickel?

  • number of coordinate bonds?
  • bond angle?
A

Platinum and Nickel complexes form in a square planar shape

-this consists of four coordinate binds with a bond angle of 90 degrees

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51
Q

what is the shape and bond angle of silver complexes?

A
  • silver complexes always have a linear shape with 2 coordinate bonds around the central metal ion.
  • the bond angle is 180 degrees
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52
Q

what is the type of complex present in Tollen’s reagent?

A

linear, silver complexes.

[Ag(NH3)2]+

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53
Q

what is cis platin?

A
  • this is the cis isomer of a square planar complex of platinum.
  • it is commonly used a a cancer therapy drug
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54
Q

draw the structure of cis platin

A

insert pic page 8
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Notes/AQA/Inorganic-II/Detailed/2.5.%20Transition%20Metals.pdf

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55
Q

why is the enthalpy change for ligand substitution near zero?

A

the enthalpy change for ligand substitution is very small as the bonds being formed are very similar to the bonds that were broken.

-therefore overall enthalpy change is near zero

56
Q

draw the cis and trans isomers for nicl2(nh3)2

A

page 4

https://chemrevise.files.wordpress.com/2022/01/2.5-revision-guide-transition-metals-aqa.pdf

57
Q

draw the cis and trans isomers for [Cr (H2O)4 Cl2]+

A

page 4

https://chemrevise.files.wordpress.com/2022/01/2.5-revision-guide-transition-metals-aqa.pdf

58
Q

what factors does colour changes arise from?

A

changes in:

  1. oxidation state,
  2. co-ordination number
  3. ligand.
59
Q

what is the general structure of naming complexes?

A

prefix-ligand-metal

60
Q

what is the ligand name for Cl-

A

chloro

61
Q

what is the ligand name for H2O?

A

aqua

62
Q

what is the ligand name for NH3?

A

ammine

63
Q

what is the ligand name for OH-?

A

hydroxo

64
Q

what is the ligand name for CN-?

A

cyano

65
Q

what is the difference between copper (II) and cuprate (II)?

A

copper (II)= 2 copper

cuprate (II) = 4 copper

66
Q

what is the difference between silver (I) and argentate (I)?

A

silver (I) is positively charged and argentate is negatively charged

67
Q

what is the difference between cobalt (I) and cobaltate (II)?

A

cobalt (I) is positively charged and cobaltate (II) is negatively charged

68
Q

what is the difference between iron (II or Iron (III) and ferrate (II) or ferrate (III)?

A

iron (II and Iron (III) are positive and ferrate (II) or ferrate (III) are negative

69
Q

write an equation where only the oxidation state change of [Co(NH3)6]2+ results in a colour change and what is the colour change?

A

O2
[Co(NH3)6] 2+ (aq) -> [Co(NH3)6] 3+ (aq) + e-
yellow brown

70
Q

write an equation where both the ligand and co-ordination number are changing resulting a colour change of [Co(H2O)6] 2+ and what is the colour change?

A

[Co(H2O)6] 2+ + 4Cl- -> [CoCl4] 2- + 6H2O

pink blue

71
Q

write an equation where only the ligand number of [Co(H2O)6] 2+ changes resulting in a colour change and what is the colour change?

A

[Co(H2O)6] 2+ +6NH3 -> [Co(NH3)6] 2+ + 6H2O

pink yellow brown

72
Q

how does a colour chnage arise in transition metals?

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

73
Q

what are the equations that link the colour, wavelength and frequency of the light absorbed with the energy difference between the split d orbitals.

A 
ΔE = hv
ΔE = hc / λ
v = frequency of light absorbed (unit s-1 or Hz)
h = Planck's constant 6.63 x 10^-34
ΔE = energy difference between spilt orbitals (J) 
c = speed of light 3.00 x 10 ^8 (m s-1)
λ = wavelength of light absorbed (m)
74
Q

what does changing a ligand or changing the coordination number lead to?

A

it will alter the energy split between the d- orbitals. changing ΔE and hence the frequency of light absorbed

75
Q

with Scandium (Sc3+), i sthere ot an energy transfer equal to that of visible light?

A
  • 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
76
Q

with Zn2+ ions and Cu + ions, is there an energy transfer equal to that of visible light?

A

-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.

77
Q

what happens if visible light of increasing frequency is passed through a sample of a coloured complex ion?

A
  • some of the light is absorbed
  • the amount of light absorbed is proportional to the conc of the absorbing species (and to the distance travelled through the solution).
78
Q

what is the problem with complexes with only pale colours and how do we deal with this?

A
  • some complexes have only pale colours and do not absorb light strongly.
  • in these cases a suitable ligand is added to intensify the colour
79
Q

what is absorption of visible light used for in spectrometry?

A

to determine the conc of coloured ions

80
Q

describe the 5 step method for spectrophotometry

A

1- add an appropriate ligand to intensify colour
2- make up solutions of known conc
3- measure absorption or transmission
4- plot graph of absorption vs conc
5- measure absorption of unknown and compare

81
Q

why is the reasoning behind the colour of the filter in a spectrometer?

A

the colour of the filter is chosen to allow the wavelengths of light through that would be most strongly absorbed by the coloured solution

82
Q

what happens to the electron configuration of transition metals when they form ions?

A

they lose the 4s electrons before the 3d

83
Q

what are some general trends of the oxidation states of transition metals ?

A
  • Relative stability of +2 state with respect to +3 state increases across the period
  • Compounds with high oxidation states tend to be oxidising agents e.g MnO4-

•Compounds with low oxidation states are often reducing agents e.g
V2+ & Fe2+

84
Q

what is the trend in redox potential of transition metals and what is it influenced by?

A

-The redox potential for a transition metal ion changing from a higher to a lower oxidation state is influenced by pH and by the ligand.

85
Q

what are the 4 main oxidation states of Vanadium and what colour are they?

A

VO2^+ Oxidation state +5 ( a yellow solution)
VO^2+ Oxidation state + 4 (a blue solution)
V3+ Oxidation state + 3 (a green solution)
V2+ Oxidation state + 2 (a violet solution)

86
Q

what does the addition of zinc to vanadium (V) in acidic solution do?

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 would successively change from yellow to blue to green to violet

87
Q

state the properties of the ion with the V at oxidation state of +5

A
  • the ion with the V at oxidation state =5 exists as a solid compound in the form of a VO3- ion, usually as NH4VO3 (ammonium vanadate( V).
  • addition of acid to the solid will turn into the yellow solution containing the VO2^+ ion
88
Q

what does zinc metal with acid do?

A

zinc metal with acid is a strong reducing agent that can reduce most transition metal ions from a higher oxidation state to the lowest state

-e.g. Fe3+ to Fe2+

89
Q

what is [Ag(NH3)2]+ used for ?

A

[Ag(NH3)2]+ is used in Tollen’s reagent to distinguish between aldehydes and ketones.

Aldehydes reduce the silver in the Tollens’ reagent to silver

90
Q

write both the reduction half equation for [Ag(NH3)2]+

A

[Ag(NH3)2]+ + e- -> Ag + 2NH3

91
Q

in the manganate redox titration between fe2+ with MnO4-, what makes it hard to see he bottom of the meniscus in the burette?

A

the purple colour of manganate

92
Q

if the manganate is in the burette what will be the end point of the titration?

A

it will be the first permanent pink colour

93
Q

write the equation for the redox titration of manganate redox titration?

A

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

Purple (MnO4-) to colourless Mn2+

94
Q

what is the importance of choosing the correct acid for manganate titrations?

A

The acid is needed to supply the 8H+ ions. Some acids are not suitable as they set up alternative redox reactions and hence make the titration readings inaccurate.

95
Q

what acid should be used for manganate titrations?

A

only use dilute sulfuric acid for manganate titrations

96
Q

what is the importance of using the correct volumes of sulfuric acid and also using a weak acid ?

A

Insufficient volumes of sulfuric acid will mean the solution is not acidic enough and MnO2 will be produced instead of Mn2+ 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.
  • using a weak acid like ethanoic acid would have the same effect as it cannot supply the large amount of hydrogen ions needed (8H+)
97
Q

why can conc HCl not be used manganate titrations?

A
  • the Cl- ions would be oxidised to Cl2 by MnO4- due to the difference in E^0 values
  • This would lead to a greater volume of manganate being used and poisonous Cl2 being produced.
98
Q

why can nitric acid not be used for the manganate titrations?

A
  • it’s an oxidising agent meaning it would oxidise Fe2+ to Fe3+ due to the difference in E0 values
  • This would lead to a smaller volume of manganate being used.
99
Q

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 0.02 mol dm-3. 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+

A

Step1 : find moles of KMnO4 moles = conc x vol

0.02 x 9.8/1000
= 1.96x10-4 mol

Step 2 : using balanced equation find moles Fe2+ in 10cm3
= moles of KMnO4 x 5
= 9.8x10-4 mol

Step 3 : find moles Fe2+ in 100cm3 = 9.8x10-4 mol x 10

= 9.8x10-3 mol

Step 4 : find mass of Fe in 9.8x10-3 mol

mass= moles x Ar = 9.8x10-3 x 55.8 = 0.547g

Step 5 : find % mass
%mass = 0.547/2.41 x100

= 22.6%

100
Q

other than sulfuric acid, what are some other useful manganate titrations?
+
write the redox equations as well as the overall equation

A

-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

  • with ethandioate
  • Ox: C2O42-  2CO2 + 2e-

Red: MnO4-(aq) + 8H+(aq) + 5e- –> Mn2+ (aq) + 4H2O

Overall: 2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq) –> 10CO2(g) + 2Mn2+(aq) + 8H2O(l)

101
Q

as the reaction between MnO4- and C2O4^2- is slow to begin with, what do we do to speed up this reaction?
+ why is it slow?

A

-its slow because the reaction is between 2 negative ions and to do as a titration the conical flask can be heated to 60 degrees to speed up the initial reaction

102
Q

what is the ratio of MnO4- to FeC2O4 when iron (II) ethandioate reacts with both Fe2+ and the C2O4^2-?

A

1MnO4- reacts with 5Fe2+ and 2 MnO4- reacts with 5C2O42-

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

So overall the ratio is 3 MnO4- to 5 FeC2O4

103
Q

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.

A

Step1 : find moles of KMnO4
moles = conc x vol

0.0189 x 23.45/1000
= 4.43x10-4 mol

Step 2 : using balanced equation find moles FeC2O4.2H2O in 25cm3
= moles of KMnO4 x 5/3 remember the ratio 3MnO4- to 5 FeC2O4)
= 7.39x10-4 mol

Step 3 : find moles FeC2O4.2H2O in 250 cm3
= 7.39x10-4 mol x 10
= 7.39x10-3 mol

Step 4 : find mass of FeC2O4.2H2O in 7.39x10-3 mol
mass= moles x Mr = 7.39x10-3 x 179.8 = 1.33g

Step 5 : find % mass
%mass = 1.33/1.412 x100
= 94.1%

104
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.
105
Q

what are the 2 types of catalysts that transition metals and their compounds can act as?

A
  • heterogenous catalysts

- homogenous catalysts

106
Q

what is the difference between homogenous and heterogenous catalysts?

A
  • a heterogenous catalyst is in a different phase from the reactants
  • a homogenous catalyst is in the same phase as the reactants
107
Q

what is usually the state of heterogenous catalysts and the reactants in the reaction?

A

heterogenous catalysts are usually solids whereas the reactants are gaseous or in solution

-the reaction occurs at the surface of the catalyst.

108
Q

how do catalysts use the active site to speed up the rate of a reaction?

A
  • 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.
109
Q

what does the strength of adsorption tell us about catalytic activity?

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.

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

110
Q

what electrons in the electronic configuration do transition metals use to form weak bonds to the reactants?

A

they can use the 3d and 4s e- of atoms on the metal surface to form weak bonds to the reactants

111
Q

how does surface area effect catalytic activity and what do we use to speed up catalytic reactions?

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 (e.g. Rh on a ceramic support in catalytic
converters).

112
Q

what are the 4 steps in heterogenous catalysis?

A

1- reactants form bonds with atoms at active sits on the surface of the catalyst (absorbed 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)

113
Q

give examples of heterogenous catalyst and their equations:

  • catalysts in the contact process
  • catalysts used in manufacture
  • catalysts in the Haber process
A

V2O5 is used as a catalyst in the Contact process.
overall equation: 2SO2 + O2 –> 2SO3
step 1: SO2 + V2O5 –> SO3 + V2O4
step 2: 2V2O4 + O2 –> 2V2O5
* the oxidation number of the vanadium changes and then changes back. It is classed as it return s back to its original form

Cr2O3 catalyst is used in the manufacture of methanol from carbon monoxide and hydrogen.
CO + 2H2 –> CH2OH

Fe is used as a catalyst in the Haber process
N2 + 3H2 –> 2NH3

114
Q

how do catalysts become poisoned and what happens when they become poisoned?
+
give an example

A

catalysts can become poisoned by impurities and consequently have reduces 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.

115
Q

why can leaded petrol not be used in cars fitted with a catalytic coverters?

A

the lead strongly absorbs onto the surface of the catalysts

116
Q

what happens when the catalysts and reactants are in the same phase?

A
  • the reaction proceeds through an intermediate species
  • the intermediate will often have a different oxidation state to the original transition metal.
  • A the end of the reaction the original oxidation state will reoccur.
  • This illustrates importance of variable oxidation states of transition metals in catalysts.
117
Q

why are transition metals able to act as homogenous catalysts?

A
  • because they can form various oxidation states.
  • they are able to donate and receive electrons and are able to oxidize and reduce.
  • this is because the ions contain partially filled sub-shells of d electrons that can easily lose or gain electrons.
118
Q

what is the reaction between iodide and persulfate ions catalysed by + write the overall equation?

A

-the reaction between I- and S2O8^ 2- catalysed by Fe2+

Overall: S2O8^ 2- —-> 2SO4^2-

119
Q

write the separate equations for the reaction between iodide and persulfate ions and explain the reasoning between the E^0 values

A

S2O8 ^2- + 2e- —> 2SO4 ^2- E^0 +2.01V
Fe3+ + e- —> Fe2+ E^0 +0.77V
I2 + 2e- —–> 2I- E^0 +0.54V

For a substance to act as a homogenous catalyst its electrode potential must lie in between the electrode potentials of the 2 reactants. It will first reduce the reactant with the more positive electrode potential and then in the second step oxidise the reactant with the more negative electrode potential.

120
Q

write an equation for the alternative catalysed route of reaction between iodide and persulfate ions with an explanation

A

stage 1: S2O8 ^2- + 2Fe2+ —> 2SO4^ 2- + 2Fe3+
stage 2: 2I- + 2Fe3+ —> 2Fe2+ + I2

Both of the individual stages in the catalysed mechanism involve collision between positive and negative ions and will have lower activation energies

121
Q

what is the problem with using E values to find a catalyst ?

A
  • Using E values to find a catalyst only shows that catalysis is possible.
  • It does not guarantee that the rate of reaction will be increased
122
Q

why can there be another catalyst other than Fe2+ which catalyses the reaction between iodide and persulfate ions?

A

Fe3+ ions can also act as he catalyst because the 2 steps in the catalysed mechanism can occur in any order.

stage 1: S2O8 ^2- + 2Fe2+ —> 2SO4^ 2- + 2Fe3+
stage 2: 2I- + 2Fe3+ —> 2Fe2+ + I2

123
Q

what is the ion involved in the autocatalysis between ethanedioate and manganate ions?

A

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

124
Q

what is the overall equation for the autocatalytic reaction between ethanedioate and manganate ions?

A

overall; 2MnO- + 5C2o4 ^2- + 16H+ —> 2Mn2+ + 10CO2 + 8H2O

125
Q

what is the autocatalytic reaction between ethandioate and manganate ions an example of?

A

this is an example of autocatalysis where one of the products of the reaction can catalyse the reaction

126
Q

what are the 2 equations for the catalysed alternative route for the autocatalytic reaction between ethanedioate and manganate ions?

A

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

step 2: 2Mn3+ + C2O4 ^2+ 2CO2

127
Q

why is the initial uncatalysed reaction between ethanedioate and manganate ions slow + how is it sped up?

A
  • The initial uncatalysed reaction is slow because the reaction is a collision between two negative ions which repel each other leading to a high activation energy.
  • The Mn2+ ions produced act as an autocatalyst and therefore the reaction starts to speed up because they bring about the alternative reaction route with lower activation energy.
  • The reaction eventually slows as the MnO4 - concentration drops.
128
Q

how do we follow the reaction rate and evaluate it?

A
  • this can be done by removing samples at set times and titrating to work out the conc of MnO4-.
  • It could also be done by use of a spectrometer measuring the intensity of the purple colour.
  • this method has the advantages that it does not disrupt the reaction mixture, using up the reactants and it leads to a much quicker determination of conc.
129
Q

The following reaction is catalysed by CO2+ ions in an acidic solution. SO3 ^2- + 1/2 O2 —> SO4 ^2-.

Write a mechanism for the catalysed reaction by writing 2 equations involving CO2+ and CO3+ ions.

A

SO3 ^2- + 1/O2 —> SO4 ^2-
is split into these 2 reactions
=1/2 O2 + 2H+ + 2e- —> H2O
=SO3 ^2- + H2O –> SO4 ^2- + 2H+ + 2e-

add in cobalt to make 2 new redox equations. Making sure the oxidised cobalt equation is combined with the original reduced half equation and vice versa.
Co2+ —-> Co3+ + e- ——-> 1/2O2 + 2H+ + 2Co2+ –> H2O + 2Co3+

Co3+ + e- —> Co2+ ——-> 2Co3+ + SO3 ^2- + H2O –> SO4 ^2- + 2H+ + 2Co2+

ensure the 2 mechanism equations add up to the original full non- catalysed equation.

130
Q

how does silver behave like a transition metal?

A

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

131
Q

why is Silver not 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.

  • 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 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.
132
Q

what are the reactions of halides with silver nitrate? (equation + colour)

A

-fluorides produce no precipitate
-chlorides produce 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) –> Agl (s)

133
Q

write the equations where AgCl, AgBr and AgI react with ammonia

A

-silver chloride dissolves in dilute ammonia to from a complex ion
AgCl (s) + 2NH3 (aq) –> [Ag(NH3)2]+ (aq) + Cl- (aq)
colourless solution
-silver bromide dissolves in conc ammonia to from a complex ion
AgBr (s) + 2NH3 (aq) –> [Ag(NH3)2]+ (aq) + Br- (aq)
colourless solution
-silver iodide does not react with ammonia- it is too insoluble

134
Q

what is [Ag(NH3)2]+ used for and what are the redox equations?

A

[Ag(NH3)2]+ is used in Tollens’ reagent to distinguish between aldehydes and ketones.

-Aldehydes reduce the silver in the Tollens’ reagent to silver.

Red 1/2 eq: [Ag(NH3)2]+ + e- –> Ag + 2NH3
Ox 1/2 eq: CH3CHO + H2O –> CH3CO2H + 2H+ + 2e-

135
Q

how do we use silver nitrate to work out formulae of chloride containing complexes?

A
  • sometimes a compound containing a complex may have Cl- ions acting as ligands inside the complex and Cl- ions outside the complex attracted ionically to it.
  • If silver nitrate is added to such a compound it will form the silver chloride precipitate with the free chloride ions outside of the complex

AQA A level Chemistry HJ (28 decks)

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