Transition metals

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  • Created by: Hindleyc
  • Created on: 27-05-19 09:41
transition metal characteristics of elements Sc Cu arise from
an incomplete d sub-level in atoms or ions
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When forming ions lose
4s before 3d
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these characteristics include
•complex formation, •formation of coloured ions, •variable oxidation state •catalytic activity.
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Why is Zn not a transition metal?
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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complex
:is a central metal ion surrounded by ligands.
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ligand.:
An atom, ion or molecule which can donate a lone electron pair
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What bonding is involved in complex formation.
Co-ordinate bonding
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Co-ordinate bonding is when the
shared pair of electrons in the covalent bond come from only one of the bonding atoms.
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Co-ordination number:
The number of co-ordinate bonds formed to a central metal ion.
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Ligands can be
monodentate , bidentate multidentate
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monodentate
(e.g. H2O, NH3 and Cl- ) which can form one coordinate bond per ligand or
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bidentate
(e.g. NH2CH2CH2NH2 and ethanedioate ion C2O4 2- ) which have two atoms with lone pairs and can form two coordinate bonds per ligand
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multidentate
(e.g. EDTA4- which can form six coordinate bonds per ligand).
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What can can act as monodentate ligands.
H2O, NH3 and Cl−
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What are The ligands NH3 and H2O .
similar in size and are uncharged
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So Exchange of the ligands NH3 and H2O occurs
without change of co-ordination number (eg Co2+ and Cu2+).
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[Co(H2O)6]2+(aq) + 6NH3 (aq) 
[Co(NH3)6]2+(aq) + 6H2O (l)
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This substitution may, however, be
incomplete as in the case with Cu.
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Cu becomes 1
[Cu(NH3)4(H2O)2]2+ deep blue solution
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[Cu(H2O)6]2+(aq) + 4NH3 (aq)
[Cu(NH3)4(H2O)2]2+ (aq) + 4H2O (l)
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Addition of a high concentration of chloride ions (from conc HCl or saturated NaCl) to an aqueous ion leads to a
ligand substitution reaction
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but
The Cl- ligand is larger than the uncharged H2O and NH3 ligands so therefore ligand exchange can involve a change of co-ordination number.
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Addition of conc HCl to aqueous ions of Cu and Co lead to a change in.
coordination number from 6 to 4
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[CuCl4]2-
yellow/green solution
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[CoCl4]2-
blue solution
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These are
tetrahedral in shape
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[Cu(H2O)6]2+ + 4Cl-
[CuCl4]2- + 6H2O
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[Co(H2O)6]2+ + 4Cl-
[CoCl4]2- + 6H2O
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[Fe(H2O)6]3+ + 4Cl-
 [FeCl4]- + 6H2O
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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
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Ligands can be
bidentate (e.g. NH2CH2CH2NH2 and ethanedioate ion C2O4 2- ) which have two atoms with lone pairs and can form two coordinate bonds per ligand
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Ethane-1-2-diamine is a
common bidentate ligand. A complex with Ethane-1- 2-diamine bidentate ligands e.g. [Cr(NH2CH2CH2NH2)3]3+
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There are 3 bidentate ligands in this complex each bonding in twice to the metal ion. so It has a coordination number of
6
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Ethanedioate
C2O4 2-
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Acomplexwith bidentate ethanedioate ligands e.g.
[Cr(C2O4)3]3-
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shape with bond angles
Octahedral 90o
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Equations to show formation of bidentate complexes [Cu(H2O)6]2+ + 3NH2CH2CH2NH2
 [Cu(NH2CH2CH2NH2)3]2+ + 6H2O
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[Cu(H2O)6]2+ + 3C2O42- [Cu(C2O4)3]4- + 6H2O
[Cu(C2O4)3]4- + 6H2O
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What may occur when a dilute aqueous solution containing ethanedioate ions is added to a solution containing aqueous copper(II) ions.- In this reaction
Partial substitution of ethanedioate ions . four water molecules are replaced and a new complex is formed.
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[Cu(H2O)6]2+ +2C2O42-
[Cu(C2O4)2(H2O)2]2- +4H2O
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Ligands can be
multidentate (e.g. EDTA4- which can form six coordinate bonds per ligand).
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Haem is an
iron(II) complex with a multidentate ligand.
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EDTA4- anion has
with six donor sites(4O and 2N) and forms a 1:1 complex with metal(II) ions
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Oxygen forms a
co-ordinate bond to Fe(II) in haemoglobin, enabling oxygen to be transported in the blood.
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CO is toxic to humans as
CO can from 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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Equations to show formation of mutidentate complexes [Cu(H2O)6]2+ + EDTA4-
 [Cu(EDTA)]2- + 6H2O
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The substitution of monodentate ligand with a bidentate or a multidentate ligand leads to a
more stable complex. This is called the chelate effect
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This chelate effect can be explained in terms of a
positive entropy change in these reactions as more molecules of products than reactants
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[Cu(H2O)6]2+ (aq) + EDTA4- (aq)
 [Cu (EDTA)]2- (aq) + 6H2O (l)
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The copper complex ion has changed from having
unidentate ligands to a multidentate ligand.
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In this reaction there is an
increase in the entropy because there are more moles of products than reactants (from 2 to 7), creating more disorder.
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The enthalpy change is
small as there are similar numbers of bonds in both complexes.
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Free energy ΔG will be
negative as ΔS is positive and ΔH is small.
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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 lathering.
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[Co(NH3)6]2+ + 3NH2CH2CH2NH2
 [Co(NH2CH2CH2NH2)3]2+ + 6NH3
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This reaction has an.
increase in entropy because of the increase in moles from 4 to 7 in the reaction
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ΔS is
positive.
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Its enthalpy change ΔH is
close to zero as the number of dative covalent and type (N to metal coordinate bond) are the same so the energy required to break and make bonds will be the same. Therefore Free energy ΔG will be negative and the complex formed is stable.
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The formation of the stable EDTA complex with metal ions can with the choice of suitable indicator be done in a quantitative titration.
a quantitative titration.
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[Cu(H2O)6]2+ + EDTA4-
 [Cu(EDTA)]2- + 6H2O Always the same 1:1 ratio with any metal ion
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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 r
pg 3 TM
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transition metal ions commonly form (6)
octahedral complexes with small ligands (e.g. H2O and NH3). [Co(NH3)6]2 [Cu(H2O)6]2+
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transition metal ions commonly form 4
tetrahedral complexes with larger ligands (e.g.Cl- ). [CoCl4]2-
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complexes what are also formed, 4
square planar complexes e.g. cisplatin
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Ag+ commonly forms 2
linear complexes e.g. [Ag(NH3)2]+ used as Tollen’s Reagent
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Complexes can show two types of stereoisomerism:
cis-trans isomerism and optical isomerism
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cis-trans isomerism is. a special case of
e-z isomerism
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which complexes
square planar octahedral
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Complexes with 3 bidentate ligands can form
two optical isomers (non-superimposable mirror images).
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Optical isomerism in
octahedral complexes
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Colour changes arise from changes in
1. oxidation state, 2. co-ordination number 3. ligand.
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Colour arises from
electronic transitions from the ground state to excited states: between different d orbitals.
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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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dyz Octahedral complex ion Ligands cause the
5 d orbitals to split into two energy levels.
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equation links the colour and frequency of the light absorbed with the energy difference between the split d orbitals.
. E = hv
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v =
frequency of light absorbed (unit s-1 or Hz)
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h=
Planck’s constant 6.63 × 10–34 (J s)
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E =
energy difference between split orbitals (J)
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A solution will appear blue if it
absorbs orange light. The energy split in the d orbitals E will be equal to the frequency of orange light(5 x1014 s-1) x Planck’s constant E in a blue solution = hv
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Changing a ligand or changing the coordination number will
alter the energy split between the d- orbitals, changing E and hence change the frequency of light absorbed. changing colour
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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.
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In the case of Zn2+ ions and Cu+ ions the
d shell is full e.g.3d10 sothereisnospaceforelectronstotransfer. Therefore there is not an energy transfer equal to that of visible light.
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If visible light of increasing frequency is passed through a sample of a coloured complex ion, some of the light is
absorbed.
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The amount of light absorbed is proportional to the
concentration of the absorbing species (and to the distance travelled through the solution).
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Some complexes have only so?
pale colours and do not absorb light strongly. In these cases a suitable ligand is added to intensify the colour.
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Absorption of visible light is used in
spectrometry to determine the concentration of coloured ions.
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method
Add an appropriate ligand to intensify colour •Make up solutions of known concentration •Measure absorption or transmission •Plot graph of absorption vs concentration •Measure absorption of unknown and compare
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Spectrometers contain a , how chosen?
coloured filter. The colour of the filter is chosen to allow the wavelengths of light through that would be most strongly absorbed by the coloured solution.
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Transition elements show
variable oxidation states.
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General trends
•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+
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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.
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Vanadium has four main oxidation states
VO2^+ Oxidation state +5 ( a yellow solution) VO^2+ Oxidation state + 4 (a blue solution) V^3+ Oxidation state + 3 (a green solution) V^2+ Oxidation state + 2 (a violet solution)
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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)
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It is a reasonably strong
oxidising agent.- Addition of acid to the solid will turn into the yellow solution containing the VO2^+ ion.
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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
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[Ag(NH3)2]+ is used in
Tollen’s reagent to distinguish between aldehydes and ketones . Aldehydes reduce the silver in the Tollen’s reagent to silver.
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The redox titration between
Fe2+ with MnO4– (purple) is a very common exercise. This titration is self indicating because of the significant colour change from reactant to product.
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The purple colour of manganate can make it difficult to see the bottom of
meniscus in the burette.
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If the manganate is in the burette then the
end point of the titration will be the first permanent pink colour. Colourless  purple
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MnO4-(aq) + 8H+ (aq) + 5Fe2+ (aq)
Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq) (purple - colourless)
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Choosing correct acid for manganate titrations.?
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. Only use dilute sulfuric acid for mangganate titrations.
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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.
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Using a weak acid like ethanoic acid would
have the same effect as it cannot supply the large amount of hydrogen ions needed (8H+).
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It cannot be conc HCl as, This would lead to a
the Cl- ions would be oxidised to Cl2 by MnO4- , greater volume of manganate being used and poisonous Cl2 being produced.
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It cannot be nitric acid as it is an, This would lead to a
oxidising agent, smaller volume of manganate being used.
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Do q on page 7
:)
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The reaction between MnO4- and C2O42- is
slow to begin with (as the reaction is between two negative ions). To do as a titration the conical flask can be heated to 60o C to speed up the initial reaction.
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Other useful manganate titrations With
hydrogen peroxide and With ethanedioate pg 7!!!!!
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Catalysts
increase reaction rates without getting used up. They do this by providing an alternative route with a lower activation energy.
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Transition metals and their compounds can act as
heterogeneous and homogeneous catalysts.
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A heterogeneous catalyst is
in a different phase from the reactants
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A homogeneous catalyst is in
the same phase as the reactants
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Heterogeneous catalysts are usually
solids whereas the reactants are gaseous or in solution. The reaction occurs at the surface of the catalyst.
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Adsorption of reactants at active sites on the surface may lead to
catalytic action.
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The active site is the
place where the reactants adsorb on to the surface of the catalyst.
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This can result in the bonds within the reactant molecules becoming
weaker, or the molecules being held in a more reactive configuration.
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There will also be a
higher concentration of reactants at the solid surface so leading to a higher collision frequency.
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The strength of adsorption helps to determine
the effectiveness of the catalytic activity.
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Some metals e.g. W have and. Some metals e.g. Ag have
too strong adsorption and so the products cannot be released. too weak adsorption, and the reactants do not adsorb in high enough concentration.
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Ni and Pt have
about the right strength and are most useful as catalysts
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Transition Metals can use the
3d and 4s e- of atoms on the metal surface to form weak bonds to the reactants.
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Steps in Heterogeneous Catalysis
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
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4. This in turn
weakens bonds between product and catalyst and product leaves (desorbs).
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Increasing the surface area of a solid catalyst will
improve its effectiveness
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. 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).
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What is used as a catalyst in the Contact Process.
V2O5
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Overall equation : step1 step 2
2SO2 + O2  2SO3, SO2 +V2O5 SO3 + V2O4, 2V2O4 + O2  2V2O5
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Note the oxidation number of the vanadium
changes and then changes back. It is still classed as a catalyst as it returns to its original form.
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Cr2O3 catalyst is used in the manufacture of
methanol from carbon monoxide and hydrogen. CO + 2H2 CH3OH
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Fe is used as a catalyst in the
Haber Process N2 + 3H2 2NH3
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Catalysts can become
poisoned by impurities and consequently have reduced efficiency.
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Poisoning has a
cost implication e.g. poisoning by sulphur in the Haber Process and by lead in catalytic converters in cars means that catalysts lose their efficiency and may need to be replaced.
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It is important to ensure the
purity of the reactants if poisoning can occur.
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Leaded petrol cannot be used in cars fitted with a catalytic converter since
lead strongly adsorbs onto the surface of the catalyst.
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When catalysts and reactants are in the same phase,
the reaction proceeds through an intermediate species.
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The intermediate will often have a different
oxidation state to the original transition metal.
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At the end of the reaction the original oxidation state will
reoccur. This illustrates importance of variable oxidation states of transition metals in catalysis.
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Transition metals can act as homogeneous catalysts because
they can form various oxidation states.
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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.
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The reaction between I- and S2O82- catalysed by
Fe2+
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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
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For a substance to act as a homogenous catalyst its
electrode potential must lie in between the electrode potentials of the two reactants, so it can first reduce the reactant with the more positive electrode potential and then in the second step oxidize the reactant with the more negative electrode po
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eqn overall
S2O8^2- + 2I-  2SO4^2- +I2 pg 9
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Catalysed alternative route stage 1
S2O8^2- + 2Fe^2+  2SO4^2- + 2Fe3+
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stage2
2I- + 2Fe3+  2Fe2+ + I2
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Both of the individual stages in the catalysed mechanism involve
collision between positive and negative ions and will have lower activation energies.
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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
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Fe3+ ions can also act as the
catalyst because the two steps in the catalysed mechanism can occur in any order
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Autocatalytic Reaction between
Ethanedioate and Manganate ions
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overall
2 MnO4^- + 5 C2O4^2- + 16 H+2Mn2+ + 10 CO2 + 8 H2O
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The autocatalysis by Mn2+ in titrations of
C2O4 2- with MnO4-
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This is an example of autocatalysis
where one of the products of the reaction can catalyse the reaction.
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Catalysed alternative route Step 1Step 2 2Mn3+ + C2O42-  2Mn2+ + 2 CO2
4Mn^2+ + MnO4^- + 8 H+5Mn^3+ + 4 H2O , 2Mn^3+ + C2O4^2-  2Mn^2+ + 2 CO2
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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.
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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.
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The reaction eventually
slows as the MnO4- concentration drops. graph pg 9
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Following the reaction rate 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.
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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.
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eg pg 10
:)
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Ag+ commonly forms
linear complexes e.g. [Ag(H2O)2]+ [Ag(NH3)2]+, [Ag(S2O3)2]3- and [Ag(CN)2]- All are colourless solutions.
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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).
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Silver is unlike the transition metals in that it
does not form coloured compounds and does not have variable oxidation states.
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Silver complexes all have a
+1 oxidation state with a full 4d subshell (4d10).
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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
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It is not therefore able to do
electron transitions between d orbitals that enable coloured compounds to occur.
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flow diagram pg 11
:)
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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.
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If silver nitrate is added to such a compound it will
only form the silver chloride precipitate with the free chloride ions outside of the complex.
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eg on pg 11
DONE :))))
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