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Chemistry IAL EDEXCEL A-Level > Unit 5- Topic 17 > Flashcards

Unit 5- Topic 17 Flashcards

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1
Q

What are transition metals

A

Transition metals are d-block elements that form one or more stable ions with incompletely filled d-orbitals

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2
Q

Characteristics of transition metals

A

-hard solids
-high melting and boiling points
-can act as catalysts
-form coloured ions and compounds
-form ions with different oxidation numbers
-form ions with incompletely filled d-orbitals
-aqueous solutions of transition metal ions are usually coloured

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3
Q

In transition metal ions, what is the order in which the electrons fill in spdf notation

A

1s2s2p3s3p4s3d

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4
Q

Order in which electrons are lost in transition metals

A

4s3d3p3s2p2s1s

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5
Q

Why are electrons lost first in the 4s orbital rather than 3d

A

Because the 4s electrons have higher energy than the electrons in the 3d orbital.

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6
Q

What elements are the exception to the rule of electronic configuration

A

Chromium since it has 5 electrons in its 3d orbital but only 1 in its 4s orbital
Copper since it has 10 electrons in its 3d orbital but only 1 in its 4s orbital

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7
Q

How are oxidation numbers created

A

Because each element can lose a variable number of electrons to form ions with different oxidation numbers.

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8
Q

What is special about transition metal ions with high oxidation numbers

A

That they usually contain an electronegative element

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9
Q

What affects the frequency in which we see high oxidation numbers in metal ions

A

Increasing the nuclear charge means that the electrons are attracted more strongly and are less likely to be involved in bonding. So higher oxidation numbers are lesss common

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10
Q

Why do transition metals have variable oxidation states

A

Because the 4s and 3d orbitals are very close in energy levels. This not only makes it possible for electrons to be lost from both orbitals relatively easily, but also means the remaining electrons form a stable configuration.

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11
Q

What is a ligand

A

A species that uses a lone pair of electrons to form dative bond with a metal ion

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12
Q

Why can transition metals form complex ions

A

Because they have a relatively small ionic radii that enables them to attract electron-rich species more strongly

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13
Q

What should the diagram of a complex ion show

A

-Bonds are shown with lines, indicating that they are dative (coordinate) bonds - one of the lone pairs of electrons on one atom of each molecule is used to form the bond. Both electrons come from the ligand.
-whole structure is shown inside square brackets,and the original charge of the central metal ion is shown outside the brackets
-the ligand molecules are arranged in a regular pattern around the Fe- due to the electron pair repulsion theory
-solid wedges represent bonds coming out of the place of the paper
-stripped wedges represent bonds going behind the plane of the paper

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14
Q

What is the coordinaton number

A

The number of dative (coordinate) bonds in the complex

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15
Q

What is a complex ions

A

A species containing a metal ion joined to ligands with an overall positive or negative charge

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16
Q

State the name and charge of this ligand -> water

A

Name in complex- aqua
Charge- 0

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17
Q

State the name and charge of this ligand -> hydroxide

A

Name in complex- hydroxo
Charge- -1

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18
Q

State the name and charge of this ligand -> ammonia

A

Name in complex- ammine
Charge- 0

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19
Q

State the name and charge of this ligand -> chloride

A

Name in complex- chloro
Charge- -1

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20
Q

Order followed in the naming of complex ions

A

-number of ligands (eg. Hexa)
-name of ligand (s), if many order in alphabetical order (eg. Aqua)
-name of metal ion (eg. Iron or ferrate if overall negative charge of complex ion)
-oxidation number of the metal ion

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21
Q

what is a monodentate ligand

A

ligands which form one coordinate bond with the central metal ion. Eg: H2O

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22
Q

what is a bidentate ligand

A

ligands which form two coordinate bonds with the central metal ion. Eg: NH2CH2CH2NH2

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23
Q

what is a hexadentate ligand

A

ligands which form six coordinate bonds with the central metal ion. Eg: EDTA

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24
Q

what is a polydentate or multidentate ligand

A

general term for a ligand which forms more than one coordinate bond with the central metal ion

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25
Q

what are complementary colours

A

colours opposite each other on the colour wheel

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26
Q

why are complementary colours involved in the colouration of aqueous transition metal solutions

A

Colour arises because of how substances absorb and reflect light so when white light is passed through a solution containing a transition metal complex, some wavelengths of light are absorbed by the complex. The remaining wavelengths are reflected and transmitted to the human eye. The light emerging will therefore contain proportionately more of the complementary colour

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27
Q

why is aqueous solution of Zn 2+ colourless and Cu 2+ blue?

A

because ions that have a completely filled 3d energy levels (Zn)and ions that have no electrons in their 3d energy levels are not coloured (Sc). Cu 2+ has only 9 electrons in the 3d energy level so it is not completely filled.

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28
Q

why are aqueous solutions of transition metals coloured referring to electrons

A

because they have unfilled 3d energy level. So when the ligans are attached, the energy level splits into two levels with slightly different energies (from 10 spaces to 6 lower energy spaces and 4 higher energy spaces). If one of the lectrons in the lower energy level absorbs energy from visible spectrum, it will move to the higher energy level (promotion).

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29
Q

what does the amount of energy the electron absorbs at promotion depend on

A

the difference in energy between the two levels. The bigger the difference, the more energy the electron absorbs. The amount of energy the electron absorbs is directly proportional to the frequency of the absorbed light. The energy gained increases as frequency increases.

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30
Q

Relationship between energy gained by the electron in transition metal solution and the wavelength of light

A

as energy gained increases, wavelength decreases. A long wavelength is equal to a high wavelength

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31
Q

what does the colour of the transition metal solution depend on

A

the coordination number of the complex
the type of ligand bonded to the ion
the oxidation state

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32
Q

what reactions can cause the colour of a transition metal solution to change

A

redox- changes oxidation state
deprotonation- changes type of ligand bonded by gaining or losing a hydrogen
ligand exchange- changes type of ligand bonded by replacing
coordination number change

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33
Q

examples of monodentate ligands

A

H2O, OH-, NH3

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34
Q

when does an octahedral complex occur

A

when there is a six-fold coordination with monodentate ligands
bond angle: 90º
examples: metal ion with H2O, OH-, NH3

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35
Q

when does a tetrahedral complex occur

A

metal ions may form tetrahedral complexes with relatively large ions such as Cl-. due to he large size of the ligand, there is not enough room around the central metal ion for six chloride ions to act as ligands
tetrahedral complexes may present optical isomerism
bond angle: 109.5

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36
Q

when does a linear complex occur

A

a complex with only two ligands. we will only see [H3N - Ag - NH3]+

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37
Q

what is the shape of cis- and trans-platin

A

square planar complex

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38
Q

what two metals form square planar complexes

A

platinum and nickel

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39
Q

what does cis- and trans-platin consist of

A

a platinum (II) ion
two ammonia ligands
two chloride ligands

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40
Q

what does the cis- prefix indicate

A

(Z) - indicates that identical ligands are next to each other

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41
Q

what does the trans- prefix indicate

A

(E)- indicates that identical ligands are opposite each other

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42
Q

what does a square planar complex consist of

A

central metal ion has four coordinate bonds
bond angle: 90º

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43
Q

what is cis-platin used for

A

cis- platin is used for effective treatment for cancer. it kills cancer cells, because all cells contain DNA. during cell division, DNA must separate from each other to form more DNA, cis-platin enables it to form a bond between the two strands of DNA which prevents the cancer cells from dividing

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44
Q

why isn’t trans-platin used for cancer treatment

A

cis-platin is supplied as a single isomer because trans-platin and cis-platin have diiferent structures. this difference makes trans-platin less effective in cancer treatment and more toxic.

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45
Q

what is the role of haemoglobin

A

a protein in red blood cells that transports oxygen through the bloodstream in humans and other mammals

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46
Q

what is found within the protein part of haemoglobin

A

4 haem groups made up of mostly carbon and hydrogen atoms. inside each haem group, there are 4 nitrogen atoms that hold an Fe 2+ ion by forming dative bonds with it in a square planar structure. there is a fifth dative bond from the protein to the Fe 2+ ion.

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47
Q

how does haemoglobin collect oxygen when it passes through the lungs

A

the oxygen molecule acts as a ligand by using one of its lone pairs of electrons to form a dative bond with one of the Fe 2+ ions inside the haem group

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48
Q

why is carbon monoxide dangerous

A

because it contains a lone pair of electrons therefore it can act as a ligand. a ligand substitution reaction is very likely to happen since the dative bond between haemoglobin and oxygen is not very strong whereas the dative bond between carbon monoxide and haemoglobin is particularly strong

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49
Q

why is the dative bond between oxygen and haemoglobin not very strong

A

so that it is easily released when needed

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50
Q

why is the substitution of oxygen with carbon monoxide in haemoglobin dangerous

A

because one carboxyhaemoglobin is formed, the dative bond is so strong that it is not easily broken down. this means that if enough haemoglobin has converted to carboxyhaemoglobin, there may be too little oxygen to support life

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51
Q

colour of vanadium(II) or (V 2+) in aqueous solution

A

purple

52
Q

colour of vanadium(III) or (V 3+) in aqueous solution

A

green

53
Q

colour of oxovanadium(IV) or (VO 2+) in aqueous solution

A

blue

54
Q

colour of dioxovanadium(V) or (VO2 +) in aqueous solution

A

yellow

55
Q

what conditions/reactants do you need for the reduction of VO2 + to V 2+

A

an acidic solution with VO2 + and zinc begins the reduction and allows us to observe a gradual colour change

56
Q

why does zinc only reduce dioxovanadium (V) to vanadium (II)

A

because the reduction from 5+ to 4+, 4+ to 3+ and 3+ to 2+ all have a standart electrode potential which is more positive that the one of Zn, this means that Zn can release electrons for the reduction to occur as the reaction is thermodynamically feasible. in the reduction of 2+ to 0, the Eº of Zn is more positive therefore it cant release electrons and the reaction is not thermodynamically feasible

57
Q

CHROMIUM(III) WITH ALKALIS
reaction between hexaaquachromium (III) + aqueous sodium hydroxide ->

A

green solution makes a green precipitate of [Cr(H2O)3(OH)3]
equation: [Cr(H2O)6] 3+ + 3OH - -> [Cr(H2O)3(OH)3] + 3H2O
type of reaction: deprotonation, 3 hydroxide ions have removed hydrogen ions from three of the water ligands and converted them into water molecules. the three water ligands that have lost hydrogen ions are now hydroxide ligands

58
Q

CHROMIUM(III) WITH ALKALIS
reaction between hexaaquachromium (III) + aqueous ammonia ->

A

green solution makes a green precipitate of [Cr(H2O)3(OH)3]
equation: [Cr(H2O)6] 3+ + 3NH3 - -> [Cr(H2O)3(OH)3] + 3NH4 +
type of reaction: deprotonation, 3 ammonia ions have removed hydrogen ions from three of the water ligands and converted them into ammonium ions. the three water ligands that have lost hydrogen ions are now hydroxide ligands

59
Q

CHROMIUM(III) WITH ALKALIS
reaction between hexaaquachromium (III) + excess aqueous sodium hydroxide ->

A

green precipitate dissolves to form a green solution
equation: [Cr(H2O)3(OH)3] + OH- ->[Cr(H2O)2(OH)4]- + H2O
if the NaOH (aq) is more concentrated, further deprotonation occurs
equation if NaOH is more concentrated: [Cr(H2O)2(OH)4]- + 2OH- -> [Cr(OH)6] 3- + 2H2O (there is no further change in colour

60
Q

how can reactions involving hydroxide ions be reversed

A

by the addition of acid, illustrating the amphoteric nature of the neutral complex of chromium

61
Q

CHROMIUM(III) WITH ALKALIS
reaction between hexaaquachromium (III) + excess aqueous ammonia ->

A

green precipitate turns into a violet/purple solution
the ppt is slow to dissolve
equation:[Cr(H2O)3(OH)3] + 6NH3 -> [Cr(NH3)6] 3+ + 3H2O + 3OH-

62
Q

oxidation of chromium using hydrogen peroxide after the use of excess alkali

A

solution changes from green to yellow
equation: 2[Cr(OH)6] 3- +3H2O2 -> 2CrO4 2- + 2OH- +8H2O
the complex of chromate (VI) is not enclosed in square brackets

63
Q

what is the colour of [Cr(NH3)6] 3+ as a solid and why

A

as a solid it is violet. when it is dissolved it turns green because of ligand exchange reactions that occur with the negative ions present

64
Q

Why can you change from chromate (VI) ions to dichromate (IV) ions by adding acid to the first ions

A

Because chromate (VI) ions are stable in alkaline solution but in acidic conditions, dichromate (VI) ions are more stable. If acid added there is a colour change from yellow to orange.
Equation: 2CrO4 2- + 2H + <—> Cr2O7 2- + H2O
This reaction is easily reversed by adding alkali

65
Q

Equation for reduction of dichromate (VI) ions

A

When zinc is added to an ACIDIC solution with dichromate (IV) ions, reduction reaction occurs. Oxidation state changes from +6 to +3 and then to +2.
+6 to +3: colour change from orange to green
Equation: Cr2O7 2- + 14H + + 3Zn —> 2Cr 3+ + 7H2O + 3Zn 2+

+3 to +2: colour change from green to blue
Equation: 2Cr 3+ + Zn —> 2Cr 2+ + Zn 2+

66
Q

Equation for the OXIDATION of chromium (III) to chromium (VI)

A

Oxidation - loss of electrons, the species has to be electron releasing
Oxidation of chromium (III) by hydrogen peroxide in alkaline condition:
2Cr(OH)3 + 3H2O2 + 4OH- —> 2CrO4 2- + 8H2O

If the solution is then acidified by adding dilute sulfuric acid, the chromate (VI) is converted into dichromate (VI)
Equation:: 2CrO4 2- + 2H+ —> Cr2O7 2- + H2O

67
Q

Equation for the reduction of chromium (VI) ions to chromium (III)

A

Reduction of chromium (VI) by zinc in acidic conditions
Cr2O7 2- + 14H+ + 3Zn —> 2Cr 3+ + 7H2O + 3Zn 2+

68
Q

Reduction of chromium (III) to chromium (II)

A

Reduction of chromium (III) by zinc in acidic conditions
2Cr 3+ + Zn —> 2Cr 2+ + Zn 2+

69
Q

Colour of Fe 2+ ions in solution

A

Pale green

70
Q

What happens to an aqueous solution of Fe 2+ ions when it is exposed to air

A

Colour changes gradually from pale green to yellow/ brown as thee oxidation number of iron increases from +2 to +3. The type and number of ligands remains unchanged in this oxidation reaction.
Complexes:[Fe(H2O6] 2+ and [Fe(H2O)6] 3+

71
Q

Reaction of copper (ll) sulfate solution and aqueous sodium hydroxide

A

Pale blue solution forms a blue precipitate.
Equation: [Cu(H2O)6] 2+ + 2OH- —>[Cu(H2O)4(OH)2] + 2H2O
Reaction: deprotonation - two hydroxide ions have removed hydrogen ions from two of the water ligands and converted them into water molecules, the water ligands have lost hydrogen ions are now hydroxide ligands

72
Q

Reaction of copper (ll) sulfate solution and aqueous ammonia

A

Pale blue solution forms a blue precipitate
Equation: [Cu(H2O)6] 2+ + 2NH3 —> [Cu(H2O)4(OH)2] + 2NH4 +
Reaction: deprotonation - two of the water ligands transfer a hydrogen ion to the ammonia molecules

73
Q

Reaction of copper (ll) sulfate solution and excess aqueous ammonia

A

aqueous amonia added to the blue precipitate formed after addition of some ammonia, blue ppt dissolves to form a deep blue solution
Equation: [Cu(H2O)4(OH)2] + 4NH3 —> [Cu(NH3)4(H2O)2] 2+ + 2H2O + 2OH-
Reaction: ligand exchange - four ammonia molecules replace two water molecules and two hydroxide ions.

74
Q

what do reactions causing a change in coordination number involve

A

always involve a change of ligand

75
Q

reaction between copper (II) sulfate solution and concentrated hydrochloric acid

A

colour changes gradually from blue to green and finally to yellow
equation: [Cu(H2O)6] 2+ + 4Cl- <–> [CuCl4] 2- + 6H2O
reaction: ligand exchange- six water ligands have been replaced by four chloride ions. change in coordination number, 6 to 4. there is no change in oxidation number

76
Q

what is amphoteric behaviours

A

the ability of a species to react with both acids and bases

77
Q

what is an amphoteric substance

A

a substance that can act both as an acid and as a base

78
Q

what is common of the metal hydroxides formed in the reaction with aqueous ammonia and sodium hydroxide

A

they will all dissolve in acid and accept protons. some metal hydroxides will also react with bases and donate protons (amphoteric)

79
Q

example of the amphoteric behaviour of chromium

A

reacting with acids: [Cr(H2O)3(OH)3] + 3H+ –> [Cr(H2O)6] 3+

reacting with bases: [Cr(H2O)3(OH)3] + 3OH- –> [Cr(OH)6] 3- + 3H2O

80
Q

reaction aqueous sodium hydroxide and hexaaquamanganese (II)

A

pale pink solution forms pale brown ppt
equation: [Mn(H2O)6] 2+ + 2OH- –> [Mn(H2O)4(OH)2] + 2H2O
reaction: deprotonation - the two hydroxide ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

will have no change when excess sodium hydroxide is used

81
Q

what happens to the pale brown ppt formed in sodium hydroxide/ ammonia is added to Mn ions when it is in contact with air

A

the pale brown ppt turns darker brown on standing in air as it is oxidised to [Mn(H2O)3(OH)3], and then turns very dark brown as it forms hydrated manganese (IV) oxide, MnO2 · xH2O

82
Q

reaction excess aqueous sodium hydroxide and hexaaquamanganese (II)

A

the pale brown precipitate will not dissolve as with in excess ammonia

83
Q

reaction aqueous ammonia and hexaaquamanganese (II)

A

pale pink solution forms pale brown ppt
equation: [Mn(H2O)6] 2+ + 2NH3 –> [Mn(H2O)4(OH)2] + 2NH4 +
reaction: deprotonation - the two hydroxide ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

84
Q

what is important to note with the colouration of [Mn(H2O)6] 2+ and [Mn(H2O)4(OH)2]

A

[Mn(H2O)6] 2+ : dilute solutions containing this complex are very likely to appear colourless, not pale pink

[Mn(H2O)4(OH)2] : true colour of complex is white, but white may not be seen as ppt rapidly turns pale brown

85
Q

reaction aqueous sodium hydroxide and hexaaquacobalt (II)

A

pink solution forms blue ppt
equation: [Co(H2O)6] 2+ + 2OH- –> [Co(H2O)4(OH)2] + 2H2O
reaction: deprotonation - the two hydroxide ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

86
Q

reaction aqueous ammonia and hexaaquacobalt (II)

A

pink solution forms blue ppt
equation: [Co(H2O)6] 2+ + 2NH3 –> [Co(H2O)4(OH)2] + 2NH4 +
reaction: deprotonation - the two ammonia have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

87
Q

what happens to the blue ppt formed in sodium hydroxide/ ammonia is added to Co (II) ions when it is in contact with air

A

the blue ppt gradually changes to pink

88
Q

reaction excess aqueous ammonia and hexaaquacobalt (II)

A

blue ppt dissolves into a pale yellow solution
equation: [Co(H2O)4(OH)2] + 6NH3 –> [Co(NH3)6] 2+ + 4H2O + 2OH-
ligand exchange - six ammonia ligands used !!!

89
Q

what happens upon standing to the pale yellow solution made from excess amonia and Co ions.

A

pale yellow solution turns into a darker yellow as [Co(NH3)6] 3+ ions forms. this is because of the oxidation by oxygen in the atmosphere allowing the oxidation number of cobalt to increase from +2 to +3. the resulting solution usually looks brown because other products are formed in the ligand exchange reaction with any water and negative molecules present

90
Q

reaction between hexaaquacobalt (II) ion and concentrated hydrochloric acid

A

added slowly.
pink solution gradually changes to blue
equation: [Co(H2O)6] 2+ + 4Cl- <–> [CoCl4] 2- + 6H2O
reaction: ligand exchange- six water ligands have been replaced by four chloride ions. change in coordination number, 6 to 4. there is no change in oxidation number

91
Q

reaction aqueous sodium hydroxide and hexaaquairon (II)

A

pale green solution forms green ppt
equation: [Fe(H2O)6] 2+ + 2OH- –> [Fe(H2O)4(OH)2] + 2H2O
reaction: deprotonation - the two hydroxide ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

no change when excess sodium hydroxide

92
Q

reaction aqueous ammonia and hexaaquairon (II)

A

pale green solution forms green ppt
equation: [Fe(H2O)6] 2+ + 2NH3- –> [Fe(H2O)4(OH)2] + 2NH4 +
reaction: deprotonation - the two ammonia ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

no change when excess ammonia

93
Q

what happens upon standing to the green precipitate made from some ammonia/ sodium hydroxide and Fe 2+ ions.

A

colour of green ppt gradually changes to brown as oxygen from the atmosphere causes oxidation, forming [Fe(H2O)3(OH)3]

94
Q

reaction aqueous sodium hydroxide and hexaaquairon (III)

A

yellow brown solution forms brown ppt
equation: [Fe(H2O)6] 3+ + 3OH- –> [Fe(H2O)3(OH)3] + 3H2O
reaction: deprotonation - the three hydroxide ions have removed hydrogen ions from three of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

no change when excess sodium hydroxide, no change upon standing

95
Q

reaction aqueous ammonia and hexaaquairon (III)

A

yellow brown solution forms brown ppt
equation: [Fe(H2O)6] 3+ + 3NH3 –> [Fe(H2O)3(OH)3] + 3NH4 +
reaction: deprotonation - the three ammonia have removed hydrogen ions from three of the water ligands and converted them into ammonium ions. Water ligands which have lost hydrogens are now hydroxide ligands.

no change when excess ammonia, no change upon standing

96
Q

when does atmospheric oxygen affect the species made upon standing

A

atmospheric oxidation occurs with complexes containing a transition metal ion with a +2 oxidation number but not for those with +3

97
Q

reaction aqueous sodium hydroxide and hexaaquanickel (II)

A

green solution forms green ppt
equation: [Ni(H2O)6] 2+ + 2OH- –> [Ni(H2O)4(OH)2] + 2H2O
reaction: deprotonation - the two hydroxide ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

no change when excess sodium hydroxide

98
Q

reaction aqueous ammonia and hexaaquanickel (II)

A

green solution forms green ppt
equation: [Ni(H2O)6] 2+ + 2NH3 –> [Ni(H2O)4(OH)2] + 2NH4 +
reaction: deprotonation - the two ammonia ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

99
Q

reaction excess aqueous ammonia and hexaaquanickel (II)

A

green ppt dissolves to form deep blue solution
equation: [Ni(H2O)4(OH)2] + 6NH3 –> [Ni(NH3)6] 2+ + 4H2O + 2OH-
reaction: ligand exchange

100
Q

reaction aqueous sodium hydroxide and hexaaquazinc (II)

A

colourless solution forms white ppt
equation: [Zn(H2O)6] 2+ + 2OH- –> [Zn(H2O)4(OH)2] + 2H2O
reaction: deprotonation - the two hydroxide ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

101
Q

reaction excess aqueous sodium hydroxide and hexaaquazinc (II)

A

white ppt (from reaction with some) dissolves to form a colourless solution
equation: [Zn(H2O)4(OH)2] + 2OH- –> [Zn(H2O)2(OH)4] 2- + 2H2O
reaction: deprotonation

102
Q

reaction aqueous ammonia and hexaaquazinc (II)

A

colourless solution forms white ppt
equation: [Zn(H2O)6] 2+ + 2NH3 –> [Zn(H2O)4(OH)2] + 2NH4 +
reaction: deprotonation - the two ammonia ions have removed hydrogen ions from two of the water ligands and converted them into water molecules. Water ligands which have lost hydrogens are now hydroxide ligands.

103
Q

reaction excess aqueous ammonia and hexaaquanickel (II)

A

white ppt (from reaction with some) dissolves to form a colourless solution
equation: [Zn(H2O)4(OH)2] + 4NH3 –> [Zn(NH3)4] 2+ + 4H2O + 2OH-
reaction: ligand exchange

104
Q

equation for amphoteric behaviour of zinc hydroxide reacting with acids

A

[Zn(H2O)4(OH)2] + 2H+ –> [Zn(H2O)6] 2+

105
Q

what refers to as the stability of a complex

A

most complexes are stable - they do not decompose easily. the stability of a complex refers to the comparison of the stabilities of two complexes in which the number of ligands has changed

106
Q

why the substitution of a monodentate ligand by a bidentate/ hexadentate ligand leads to a more stable complex ion

A

eg. 1: [Cu(H2O)6] 2+ + 3en –> [Cu(en)3] 2+ + 6H2O
or 2: [Cu(H2O)6] 2+ + EDTA 4- + –> [CuEDTA)] 2- + 6H2O
total number of species has increased from four/ two to seven. so the system becomes more disordered resulting in an increase in change of entropy of system. this ligand exchange reactions lead to increase in stability of products compared to reactants so formation of products is favoured

107
Q

what can act as heterogeneous and homogeneous catalysts

A

transition metals and their compounds

108
Q

what is a heterogeneous catalyst

A

a catalyst that is in a different phase from the reactants. the reaction occurs at the surface of the catalyst, thy are normally used in a finely divided form as small particles

109
Q

Advantages of using heterogeneous catalyst rather than homogeneous

A

Heterogeneous catalyst makes it easier to separate the reaction products from the catalyst since they are in different phases

110
Q

Use for sulfuric acid

A

Used in the manufacture of fertilisers

111
Q

Describe the Contact process

A

Conversion of sulfur dioxide to sulfur trioxide:
2SO2 + O2 <—> 2SO3
-very high temperature and pressure which makes all of the substances gases.
-mixture of reactants is passed over catalyst of vanadium (V) oxide, V2O5

112
Q

How does a heterogeneous catalyst work

A

1- Adsorption, one or more reactants become attached to the surface of the catalyst
2- Reaction, following the weakening of bonds in the adsorbed reactants
3- Desorption, in which the reaction product becomes detached from the surface of the catalyst

113
Q

Steps in the contact process

A

1: sulfur dioxide adsorbs onto the vanadium (V) oxide and a redox reaction- V2O5 + SO2 —> V2O4 + SO3
The oxidation number of vanadium decreases from +5 to +4. The sulfur trioxide then desorbs
2: oxygen reacts with the V2O4 on the surface of the catalyst and another redox reactions occurs- V2O4 + 1/2O2 —> V2O5
The original catalyst is regenerated as the oxidation number increases from +4 to +5

114
Q

What is the consequence of increased road vehicle usage

A

Increased amount of pollution especially from carbon monoxide and nitrogen monoxide

115
Q

Why is carbon monoxide dangerous (as a pollutant)

A

It is a toxic gas that interferes with oxygen transport from the lungs through the bloodstream to vital organs in the body

116
Q

Why is nitrogen monoxide dangerous as a pollutant

A

Because it is easily oxidised in the atmosphere to nitrogen dioxide. It can act as a respiratory irritant and contribute to the formation of acid rain

117
Q

How is carbon monoxide formed in car engines

A

It forms through the incomplete combustion of hydrocarbon fuels

118
Q

How is nitrogen monoxide formed in car engines

A

Through the reaction between nitrogen and oxygen at the high temperatures that exist in an internal combustion engine

119
Q

How do heterogeneous catalysts help to reduce the increased pollution from car engines

A

Car engines are fitted with catalytic converters in an attempt to reduce the effect of vehicle emissions. The main transition metal used are platinum and rhodium. They work by allowing CO and NO to be adsorbed onto the surface, since their bonds are weakened, they react together to form CO2 and N2, which are then desorbed from the surface of the catalyst:
2CO + 2NO —> 2CO2 + N2

120
Q

Difference between absorption and adsorption

A

Absorption: involves one substance becoming distributed throughout another
Adsorption: only happens a the surface of a substance

121
Q

What is an homogeneous catalyst

A

A catalyst that is in the same phase as the reactants. This catalysed reaction will proceed via an intermediate species

122
Q

Why does the reaction of the peroxydisulfate ion and iodide

A

S2O8 2- ion acts as an oxidising agent in its reaction with iodide ions.
Equation: S2O8 2- + 2I - —> 2SO4 2- + I2
This reaction at room temperature since the two reactant ions are both negatively charged and so repel each other. Fe 2+ ions are used as a catalyst to speed the reaction up

123
Q

Reaction of the S2O8 2- + 2I- in the presence of Fe 2+ as a catalyst

A

Step 1: Fe 2+ ions not repelled by the S2O8 2- because the have opposite charge.
S2O8 2- + 2Fe 2+ —> 2SO4 2- + 2Fe 3+
Step 2: the Fe 3+ ions formed react with the I- ions due to the opposite charges
2Fe 3+ + 2I- —> 2Fe 2+ + I2
The iron (II) ions are used ad regenerated, so the two steps can repeat continuously

124
Q

Reaction of the S2O8 2- + 2I- in the presence of Fe 3+ as a catalyst

A

Step 1: 2Fe 3+ + 2I- —> 2Fe 2+ + I2
Step 2: S2O8 2- + 2Fe 2+ —> 2SO4 2- + 2Fe 3+

125
Q

Reaction: oxidation of ethanedioate ions by potassium manganate (VII) in titrations

A

In the titrations in which potassium manganate (VII) in acidic conditions acts as an oxidising agent, the reactions have to be fast in order for the titration to work well enough, allowing the end point to be accurately observed.
Equation: 2MnO4 - + 5C2O4 2- + 16H+ —> 2Mn 2+ + 5CO2 + 8H2O
The reacting species are both negatively charged so the reaction is slow. As more potassium manganate (VII) solution is added, the reaction rate increases. Since the reaction produces Mn 2+ ions this can act as a catalyst therefore causing autocatalysis

Chemistry IAL EDEXCEL A-Level (7 decks)

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