Transition Metals

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  • Created by: Ellie
  • Created on: 27-01-15 20:14

Ligands

D block elements - Have highest energy electrons in a d sub-level

Transition element - Form stable ions with an incomplete d subshell 

D block ions lose 4s electrons before 3d

Properties:

  • They make good catalysts
  • They have variable oxidation states 
  • They form coloured compounds 

Complex ions - A transition metal ion bonded to one or more ligands by dative covalent bonds.

Ligand - A molecule or ion that can donate a lone pair of electrons to a central metal ion to form a dative covalent bond. 

Unidentate ligand - No. of coordinate bonds are equal to them number of ligands

  • Can only have 4 Cl because Cl is a bigger molecule so can't fit as many around the metal ion. 
  • Sometimes get square planar such as Platin. Cis platin is used to treat cancer as it binds irreversibly with DNA so it can't replicate 

Bidentate ligand - A ligand that contains two atoms that donate lone pairs. 

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Precipitation Reactions

  • An example of a multidentate ligand is EDTA, the EDTA ion forms all 6 coordinate bonds. EDTA has a 4- charge.

Precipitation Reactions:

Copper Suplhate solution: (Blue solution to blue precipitate)

Cu 2+  + 2OH-   --------> Cu(OH)2

Cobalt (II) solution: (Pink solution to blue precipitate)

Co 2+  + 2OH-  --------> Co(OH)2

Iron (II) sulphate solution: (Pale green solution to green precipitate)

Fe 2+  + 2OH-  -------> Fe(OH)2

Iron (III) sulphate solution: (Brown/Orange solution to brown-orange precipitate)

Fe 3+  + 3OH-  ----------> Fe(OH)3 


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Ligand Substitution

Copper (II) solution with ammonia: (blue solution to sark blue solution)

[Cu(H2O)6]2+  + 4NH3 -------->[Cu(H2O)2(NH3)4]2+  + 4H2O

Copper (II) solution with Chloride: (blue solution to green/yellow solution)

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

Colbalt (II) chloride solution: (pink solution to blue solution)

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


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Redox

Half Equations:

  • If electrons on the left its reduction.
  • If electrons on the right its oxidation 
  • Balance oxygen by adding H2O
  • Balance H by adding H+

Adding half equations together:

  1. Balance electrons in each equation to remove them. Can't be present in the overall equation.
  2. Times everything in the equation by the number needed.
  3. If needed cancel out any H2O or H+

Ionic equations using oxidation numbers:

  1. Balance atoms
  2. Work out oxidation number changes
  3. Use the change to balance out opposite compounds
  4. Balance any O and H 
  5. Check atoms and charges balance
  6. Simplify if needed
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