(I) Topic 23 - The Transition Metals
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What are the 4 properties of transition metals?
• Variable oxidation state • Catalytic action • Coloured compounds • Formation of complexes
What are the 4 shapes that complexes can form? Which metals do these form with? What is the coordination number for each shape?
• Linear - Ag+ - coordination no = 2 • Squar Planar - Pt2+ and Ni2+ complexes - coordination no = 4 • Tetrahedral - complexes with larger ligands, e.g. Cl- - coordination no = 4 • Octahedral - most complexes with smaller ligands, e.g. H2O and NH3 - Coordination no = 6
Name two Bidentate ligands and draw them
• 1,2 diaminoethane • Ethandioate ion For diagrams refer to notes
Explain how heterogeneous catalysis works
• The molecules are adsorbed into the catalyst • Bonds weaken and break in the reactant • New bonds form • The product is desorbed from the catalyst
What happens if adsorbtion in a heterogeneous catalyst is too weak or strong?
• too weak - not many molecules are adsorbed and catalyst has little effect • too strong - molecules will not be able to move around in the active site and so is less likely to meet other reactants
Name two heterogeneous catalysis process and the catalysts
• Haber Process - Fe • Contact Process - V2O5
Write the whole and half equations for the contact process
Refer to notes for answers
Write the equation for the Haber process
Refer to notes for answers
Name two examples of homogeneous catalysis and name their catalysts
• Iodide Ions & Peroxodisulfate ions - Fe2+ or Fe3+ • Autocatalysis between manganate ions (VII) and ethandioate
Write the whole and half equations for the reaction between Iodide ions and peroxodisulfate ions
Refer to notes for answers (Reaction is slow to begin with as both ions are - charge and repel)
Write the uncatalysed whole and half equations for the autocatalysis between manganate ions (VII) and ethandioate, along with the catalysed equations
Refer to notes for answers
Why do transition metal ions and complexes appear coloured?
• When a complex ion is formed, d-orbital splitting occurs • Visible light hits the complex, and wavelengths of light are absorbed that correspond to the energy gap, ^E where ^E=hv. • This causes an electron to be excited and promoted from the ground to higher energy state • Colours that aren’t absorbed are reflected or transmitted, causing the compound to appear that colour
What is the colour of transition metal complexes affected by?
• Metal ion • Coordinate number • Ligands • Oxidation state
What are the colours of vanadium in its different occidation states?
VO2^(+) (V=+5) - Yellow VO^(2+) (V=+4) - Blue V^(3+) (V=+3) - Green V^(2+) (V=+2) - Violet
Name one multidentate ligand
EDTA (4-) coordinatley bonds 6 times
What is cis-trans isomerism
A case of E-Z isomerism that can occur in octahedral and square planar complexes where ther are two ligands of one type different to the other ligands. cis = same side trans = opposite sides
Which conditions is it easier to oxidise a transition metal in?
alkaline conditions
Which conditions is it easier to reduce a transition metal in?
Acidic
Using KMnO4 in redox titrations:
• Self indicating - MnO4- react to become colourless Mn2+, so once all reactant in flask is used up solution will remain purple • Needs to be used in acidic conditions, specifically dilute sulfuric acid
Why does dilute Sulfuric acid need to be used specifically in titrations with KMnO4
• Not HCl as MnO4- would oxidise Cl- to Cl2 affecting volume • Not conc sulfuric acid as its an oxidising agent itself • Ethanoic acid is too weak of an acid
The higher the charge on the metal in a metal aqua ion, the more acidic the solution. why?
The higher the charge on the metal ion, the stronger the pull on the elctrons away from the H towards the O in the H2O ligand, breaking the OH bond. This causes the molecule to loose and H+ through hydrolysis, making the solution more acidic.
Observations: To [Fe(H2O)6]^(2+): 1. Add some NaOH 2. Then add excess NaOH
• Green solution forms a green precipitate [Fe(H2O)4(OH)2] • no further visible reaction
Observations: To [Cu(H2O)6]^(2+): 1. Add some NaOH 2. Then add excess NaOH
• Blue solution forms a blue precipitate [Cu(H2O)4(OH)2] • No further visible reaction
Observations: To [Fe(H2O)6]^(3+): 1. Add some NaOH 2. Then add excess NaOH
• Orange-Brown solution forms brown precipitate [Fe(H2O)3(OH)3] • No further visible reaction