Transition Metals

• [Cu(H20)6]2+ + 4NH3 = [Cu(NH3)4]2+ + 6H20

  This type of reaction is usually known as a Ligand Exchange Reaction
• The type of bonding that normally occurs between the metal ion and the ligand in a complex is known as Dative covalent.
• Catalytic Converters are used in cars because they convert nitrogen monoxide to nitrogen and carbon dioxide by adsorbing the reactant on its surface so weakening their bonds.
• Pt(NH3)2Cl2 has a square planar shape of its complex AND forms a stereoisomer
• CrCl4- has a tetrahedral shape and doesn't form a stereoisomer.
• Chromium electronic configuration = [Ar]3d^5 4s^1
• The chromium 3+ ion has an electronic configuration of [Ar]3d^3 4s^0
• Cr(OH)3 forms a green precipitate when concentrated ammonia is added to chromium (III) sulfate.

  When excess is added, it dissolves into [Cr(NH3)6]3+
• A hydrated transition metal ion will be colourless, the electronic configuration could be... [Ar] 3d^10
• Chromium +6 = orange (Cr207^2-)

  Chromium + 6 = yellow (Cr04^2-)

  Chromium +3 = green

  Chromium +2 = Blue
• tetrahedral = CrCl4^2-
• Zn(H20)6^2+ = colourless as fully hydrated!
• Cu(edta)^2- has a high entropy change of reaction (more positive) due to stability of the edta group
• Total number of particles on RHS being higher than LHS will equal a shift in equilibrium and more of the product will be produced.
• transition metals:

  • They have variable oxidation states
  • They form coloured ions
  • They form complex ions in solution
• A transition metal is an element with one or more stable ions with partially filled d-orbitals.
• Sodium can be used to reduce Chromium ions complexes. Sodium can be used because it has a more negative E cell value.

  E cell = -0.41 - (-2.71) = 2.3 V

  2.3 is greater than 0.6 and therefore this is a feasible reaction as well.
• Bidentate Ligand = forms two dative covalent bonds with the metal ion per molecule
• How can something acts as a bidentate ligand?

  e.g. Two nitrogen atoms are far enough apart to be able to 'wrap' around the metal cation and form dative bonds. Whereas if the nitrogen atoms were to be too close they would not be able to do this.
• The Contact Process:

  Heterogeneous = the catalyst is in a different state to the rest of the reactants and works to weaken intramolecular bonding.
• Catalyst = a substance that increases the rate of reaction by offering an alterative reaction pathway which has a lower activation energy. The catalyst can absorb the reactants onto its surface and form relatively strong bonds to it. This weakens the bonds in the reactants by absorbing electron density from the reactants' bonds, this means less energy is required to break the bonds and start the reaction. Vanadium has the ability to occupy many different oxidation states, which allows it to oxidise many other molecules. The products desorb after the reaction ends.
• Equations for Contact Process:

  • Sulphur dioxide to sulphur trioxide

  overall: SO2 + 1/2 O2 = SO3 (V2O5 as catalyst)

  intermediate:

  • V2O5 + SO2 = 2VO2 + SO3
  • 2VO2 + 1/2 O2 = V2O5 (catalyst reformed)
• Overall equation of VO2+ being reduced to V2+ by zinc in acidic solution:

  2 VO2+ + 8 H+ + 3 Zn = 2 V2+ + 4H2O + 3 Zn 2+
• Calculating % by mass of vanadium, in a sample of ammonium vanadate (V):

  e.g.

  
  

  5V2+ + 3MnO4- + 4H+ = 5VO2+ + 3Mn2+ + 2H2O

  1. Moles Mn04- = 38cm3/1000 * 0.0200 mol dm-3 = 7.6x10-4mol
  2. Moles V2+ =7.6x10-4mol/3 and then x5 = 1.267 x 10-3 mol
  3. Mol VO2+ initially = 1.267 x 10-3 mol
  4. Mass VO2+ = 1.267 x 10-3 mol x (50.9) = 0.0644903g
  5. 0.0644903/0.160 x1000 = 40.3% (3.s.f)
• Ionic Equation for the addition of HCl to copper (II) sulphate:

  • [Cu (H2O)6] 2+ + 4Cl- = [Cu(Cl)4]2- + 6H2O
  • Solution turns from blue to yellow
  • This is a Ligand Substitution reaction, Cl - replacing H2O as ligands.
  • Only 4 Cl- = change in coordination number as Cl- ions are much bigger
• Ionic eq. for the reaction between ammonia solution and hydrated copper (II) ions to form a blue precip:

  • [CU(h2O)6]2+ + 2NH3 = [Cu(h2o)4(OH)2] + 2NH4+

  Ionic eq. for the addition of excess ammonia to previous solution to form a dark blue solution:

  • [Cu(h2o)4(OH)2] + 4NH3 = [Cu(NH3)4(h2o)2]2+ + 2H20 + 2OH-