Transition metals

A d - block element is found between Group 2 and Group 3 on the Periodic table.

 

A Transition element is a d - block element that forms at least one ion with an incomplete d - sub shell.

            Sc        1s22s22p63s23p63d14s2                       Sc3+        1s22s22p63s23p6

            Zn        1s22s22p63s23p63d104s2                      Zn2+        1s22s22p63s23p63d10

Writing electron configurations

                               

 

 

Cr:      1s22s22p63s23p63d44s2  (expected)     

Cr:      1s22s22p63s23p63d54s1  (actual)

Cu:      1s22s22p63s23p63d94s2  (expected)

Cu:      1s22s22p63s23p63d104s1  (actual)

 

            A half - filled or full d - sub shell offers more stability than a full s - sub shell.

 

The electron configurations of ions:

            4s - sub shell empties before 3d - sub shell

Example 1

                        Fe:          1s22s22p63s23p63d64s2

                        Fe3+:      1s22s22p63s23p63d5

Example 2

                        Cu:         1s22s22p63s23p63d104s1

                        Cu2+:     1s22s22p63s23p63d9

Qu 1-2  P203

 

Properties of transition metal compounds

Physical properties - general metals

Physical properties - transition metals

Variable oxidation states

  • Transition elements can exist in multiple oxidation numbers.

  • The most common is 2+ as the 4s electrons are usually the first to go.

  • Because the 4s and 3d electrons are close in energy, the 3d electrons can be easily removed as well

  • This means they can form several ions by losing different numbers of electrons, all of which are fairly stable. 

  • Changes in oxidation states of the transition metals often give rise to colour changes during the reaction.

MAKE SURE YOU CAN WORK OUT OXIDATION STATES

Recap:  Rules for assigning Oxidation Numbers

 

    1)    Ox. No. of an element = 0

    2)    Ox. No. of each atom in a compound counts separately.  Sum = 0

    3)    Ox. No. of an ionic element = charge on ion.

    4)    In a polyatomic ion (SO42-), The sum of the Ox. No.s of the atoms = charge on ion.

    5)    H = +1  except with metals (metal hydrides = -1).

    6)    Gp 7 (Halogens) = -1  (except with oxygen)

    7)    O = -2  except in peroxides  (H2O2,  O = -1)

Examples:

1) KMnO4: K Mn O x 4 = 0
    +1 ? 8- = 0
      +7      
             
2) K2Cr2O7: K x 2 Cr x 2 O x 7 = 0
    +2 2 x ? 14- = 0
      +6      

Coloured compounds:

                    No partially filled d shell = no colour.  Sc3+ is white

Qu 1 - 3  P205

Catalysis and precipitation

Transition metals as catalysts

1)    Transferring electrons easily:

2)    Providing a site to react:

Haber process:

  N2(g) + 3H2(g) D 2NH3(g)        

Contact process:

  2SO2(g) + O2(g) D 2SO3(g)        

Hydrogenation of alkenes:

           

Decomposition of hydrogen peroxide:

  2H2O2(aq) 2H2O(l) + O2(g)        

Precipitating reactions

Practical

Reaction Notes
Solution colour       Precipitate colour  

Cu2+(aq)

Pale blue

+

2OH-(aq)

Cu(OH)2(s)

Blue ppt

 

           

Co2+(aq)

Pink

+

2OH-(aq)

Co(OH)2(s)

Blue ppt

u

Turns beige in the presence of air
           

Fe2+(aq)

Green

+

2OH-(aq)

Fe(OH)2(s)

Green ppt

u

Turns brown at surface in the presence of air [O]
           

Fe3+(aq)

Yellow

+

3OH-(aq)

Fe(OH)3(s)

Rusty - brown ppt

 

           

Qu 1 - 4  P207

Transition metals and complex ions

Complex ions

  Complex ion: Central metal ion is surrounded by ligands
Ligand: Molecule / ion which donates a pair of electrons forming a coordinate bond (dative bond)
Coordinate bond: One of the bonded atoms has provided both electrons for the covalent bond

 

In the example above:

Common ligands:

 

Ligand Formula Charge
Water :OH2 0
Ammonia :NH3 0
Thiocyanate :SCN- -1
Cyanide :CN- -1
Chloride :Cl- -1
Hydroxide :OH- -1

Examples:

1)        [Ti(H2O)6]3+

No Ligands  x  charge + No Ligands  x  charge = Total charge of ligands + Total charge on ion = Total charge on complex ion
5  x  0 + - = 0 + ? = 3+

            The central metal ion must be:    Ti3+

 2)       [Co(H2O)5Cl]1+

No Ligands  x  charge + No Ligands  x  charge = Total charge of ligands + Total charge on ion = Total charge on complex ion
5  x  0 + 1  x  1- = 1- + ? = 2+

            The central metal ion must be:    Co2+

Shapes of complex ions:

Qu 1 - 3

Stereoisomerism in complex ions

What is stereoisomerism?

They have the same structural formula but with a different spatial arrangement

        Cis / Trans isomerism

        Optical isomerism

Cis / Trans isomerism

1)    Octahedral with 4 of one ligand and 2 of another ligand, [Co(NH3)4Cl2]+:

  CIS = 2 ligands are at 90o to each other   TRANS = 2 ligands are at 180o to each other  

2)    Square planar with 2 of one ligand and 2 of another ligand:

  CIS = 2 ligands are at 90o to each other   TRANS = 2 ligands are at 180o to each other  

Transition metal complexes in medicine:

  • Cis platin, [PtCl2(NH3)2], is the most effective drug against cancer.
  • It binds to the DNA of fast - growing cancer cells preventing them from replicating.
  • The cells own repair system leads to the death of the cells.
  • This is the basis for chemotherapy, it has some very unpleasant side effects.

 

  • New treatments involve carboplatin.  This has fewer side effects and lower doses can be used.

 

Qu 1 - 2  P211

Bidentate and multidentate ligands

Bidentate ligands

 

 

 

[Ni(en)3]2+

  • Coordination number = 6
  • Octahedral

Cis - trans isomerism from bidentate ligands:

                                                                       

  CIS = 2 monodentate ligands are at 90o to each other   TRANS = 2 monodentate ligands are at 180o to each other  

Hexadentate ligand:

  • It is used to bind to metal ions, which reduces the concentration of those metal ions in solution.
  • Used in detergents to soften water - binds to Ca and Mg ions.
  • Binds to metals in foods stopping those metals catalyse oxidation
  • Added to blood samples stopping it clotting
  • Used to treat mercury poisoning.
  • These processes are called chelating

EDTA4-

 

Optical isomers:

        1)  3 bidentate ligands

        2)  2 bidentate ligands / 2 monodentate ligands in the cis isomer only

        3)  Hexadentate ligand

Qu 1  P213

Ligand substitution in complex ions

Ligand substitution reactions

Practical

                    These are reactions when one ligand in a complex ion is substituted with another

1)  The reaction of aq copper (II) ions and ammonia

[Cu(H2O)6]2+(aq) + 4NH3(aq)

D

[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
Pale Blue sol       Deep Blue sol    
  • 4 of the H2O ligands are replaced with 4 NH3 ligands
  • 6 ligands arranged in a distorted octahedral arrangement due to the longer Cu - N bonds

                 

In practise:

  • When you start adding ammonia you get the distinguishable pale blue precipitate of Cu2+ ions with hydroxide ions first.

  • On further addition of ammonia the deep blue colour appears.

  • This is because aq ammonia contains hydroxide ions and these react first:

Ammonia contains:

H2O(l) + NH3(aq)

D

NH4+(aq) + OH-(aq)

    a)  Copper ions react with hydroxide ions:

Cu2+(aq)

+

2OH-(aq)

Cu(OH)2(s)

 
Pale Blue       Pale Blue ppt

 

 

    b)  Copper ions react with ammonia:

[Cu(H2O)6]2+(aq) + 4NH3(aq)

D

[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
Pale Blue sol       Deep Blue sol    

Summary:

 
  [Cu(H2O)6]2+ Cu(OH)2 [Cu(NH3)4(H2O)2]2+

2)  The reaction of aq copper (II) ions and hydrochloric acid

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

D

[CuCl4]2-(aq) + 6H2O(l)
Pale Blue sol       Yellow sol    
  • 6 H2O ligands are replaced with 4 Cl- ligands.
  • This is because the Cl- ligand is larger than the H2O ligand.
  • The Cl- ligand also offers more repulsion.

                 

  • The [CuCl4]2- complex will have a tetrahedral shape around the central ion.
  • The equilibrium can be reversed by the addition of water.
  • The green intermediate is due to the mixing of the colours blue and yellow colours.

3)  The reaction of cobalt (II) ions and conc hydrochloric acid

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

D

[CoCl4]2-(aq) + 6H2O(l)
Pink sol       Dark blue sol    
  • 6 H2O ligands are replaced with 4 Cl- ligands.
  • Concentrated HCl is used to provide a high concentration of Cl- ions / ligands

                 

  • The equilibrium can be reversed by the addition of water.

Qu 1-3 P215

Ligand substitution and stability constants

Haemoglobin and ligand substitution

Haemoglobin - is made from 2 parts

a)  Haem:                                                                                            b)  Globin:

  • Haem is a non protein molecule complex that can bind to Fe2+ ions.
  • It forms 4 dative covalent bonds to Fe2+.
  • Its coordination number is 4 and binds in a square planar arrangement.
  • This allows 2 further dative covalent bonds - 1 above and 1 below.

 

  • Globin is made from 4 proteins joined together in a polypeptide.
  • They are optical isomers of each other.
  • Each protein forms a dative covalent bond to the Fe2+
  • This leaves 1 area free for a dative covalent bond.

A simpler picture:

  • Looking at one of the haem complexes which forms a further dative covalent bond with one of the proteins in globin.
  • A space is left for a further dative covalent bond.  This is where oxygen forms a dative covalent bond in order to be transported around the body.
  • It also forms a complex with carbon dioxide which is transported back to the lungs.

Video

Carbon monoxide - the silent killer

Stability constants

 

Kc

=

[PRODUCTS]p
[REACTANTS]r

Example:

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

D

[CoCl4]2-(aq) + 6H2O(l)
Kc

=

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

=

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

=

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

The value of Kstab:

Example:

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

D

[CuCl4]2-(aq) + 6H2O(l)

           [ [Cu(H2O)6]2+ ]  =  1.17  x  10-5  mol dm-3  and  [ Cl-]  =  0.08  mol dm-3   and Kstab = 4.17  x  10-5  mol-4 dm12

Kstab

=

           [ [CuCl4]2- ]           
[ [Cu(H2O)6]2+ ]   [ [Cl-] ]4
4.17  x  10-5

=

           [ [CuCl4]2- ]           
1.17  x  10-5  x  0.08

        [ [CuCl4]2- ] =  2.00 mol dm-3

Qu 1 - 3  P217

Redox titrations

Oxidation and reduction in transition metal chemistry

MnO4-  and Fe2+

Fe2+(aq)

Fe3+(aq) + e- 1
MnO4-(aq) + 8H+(aq) + 5e-

Mn2+(aq) + 4H2O(l) 2
MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)

Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Purple   Colourless

Carrying out redox titrations

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)

Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Purple   Colourless

Worked example:

25.0cm3 of a solution of iron (II) salt required 23.00cm3 of 0.0200 mol dm-3 potassium manganate (VII) for complete oxidation in acidic solution.

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)

Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
23cm3 1 : 5 25cm3          
0.02 mol dm-3 Conc = Mol x 1000/v        

= 2.3 x 10-3 x 1000/25

     
= 0.092 mol dm-3          
         
Moles = C x V/1000          
= 0.02 x 23/1000          
= 4.6 x 10-4 x 5 = 2.3 x 10-3          

Qu 1 - 2  P 219

Examples of redox titrations

    1)  Calculate the Mr and formula of an iron (II) salt

    2)  Calculating the % mass of iron in iron tablets

    3)  Calculating the purity of iron samples

    4)  Applying your knowledge to unfamiliar redox reactions / titrations

    1)  Calculate the Mr and formula of an iron (II) salt

2.950g of hydrated iron (II) sulphate, FeSO4.xH2O, was dissolved in 50cm3 of sulphuric acid.  This was made up to 250cm3 with distilled water.

25cm3 of this was titrated with 0.01 mol dm-3 KMnO4 and 21.20cm3of this was used

 

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)

Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
21.20cm3       25cm3        
0.01 mol dm-3      

(misprint in book)

       
               
               
                 
Moles = C x V/1000   1 : 5            
= 0.01 x 21.20/1000          
= 2.12 x 10-4   x 5   = 1.06 x 10-3          

    2)  Calculating the % mass of iron in iron tablets

A multivitamin tablet has a mass of 0.325g and contains iron.  The powdered tablet was dissolved in some water and sulphuric acid.

12.10cm3 of 0.002 mol dm-3 KMnO4 was titrated until the first permanent pink colour.

 

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq)

Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
12.10cm3       Mass = moles x Ar      
0.002 mol dm-3      

= 1.21 x 10-4 x 55.8

   
      = 0.00675g        
             
               
Moles = C x V/1000   1 : 5            
= 0.002 x 12.10/1000          
= 2.42 x 10-5   x 5   = 1.21 x 10-4          

Practical

    4)  Applying your knowledge to unfamiliar redox reactions / titrations

25cm3 portion of H2O2 was made up to 250cm3 with distilled water.

25cm3 portions of this was acidified and titrated with 0.0200 mol dm-3 KMnO4, 38.00cm3 of this was required to completely oxidise the hydrogen peroxide.

Calculate the original concentration of the hydrogen peroxide.  The following equation represents the oxidation of hydrogen peroxide

 

H2O2(aq)

O2(g) + 2H+ + 2e- 1
MnO4-(aq) + 8H+(aq) + 5e-

Mn2+(aq) + 4H2O(l) 2
2MnO4-(aq) + 16H+(aq) + 5H2O2(aq)

2Mn2+(aq) + 8H2O(l) + 5O2(aq) + 10H+(aq)
2MnO4-(aq) + 6H+(aq) + 5H2O2(aq)

2Mn2+(aq) + 8H2O(l) + 5O2(aq)  
Purple   Colourless

25cm3 portion of H2O2 was made up to 250cm3 with distilled water.

25cm3 portions of this was acidified and titrated with 0.0200 mol dm-3 KMnO4, 38.00cm3 of this was required to completely oxidise the hydrogen peroxide.

Calculate the original concentration of the hydrogen peroxide.  The following equation represents the oxidation of hydrogen peroxide

 

2MnO4-(aq) + 6H+(aq) + 5H2O2(aq)

2Mn2+(aq) + 8H2O(l) + 5O2(aq)
38.00cm3       25cm3        
0.02 mol dm-3      

C = moles  x  1000 / V

     
      =  1.9 x 10-3 x 1000 / 25      

= 0.0760 mol dm-3

             
               
Moles = C x V/1000   2 : 5            
= 0.02 x 38.00/1000          
= 7.6 x 10-4   x 2.5  

= 1.9 x 10-3

       

Qu 1 - 2  P221

Redox titrations - iodine and thiosulphate

2S2O32-(aq) + I2(aq)

2I-(aq) + S4O62-(aq)
    BROWN   COLOURLESS    

            1)  Iodine is liberated from it ions - using an oxidising agent (redox reaction): 

                                                                            2I-                I2    +    2e-

                        a)  Copper (II):  Cu2+

                        b)  Dichromate:  Cr2O72-

                        c)  Chlorate:  ClO-

 

            2)  The liberated iodine is titrated with a known concentration of thiosulphate,

 

2S2O32-(aq) + I2(aq)

2I-(aq) + S4O62-(aq)
    BROWN   COLOURLESS    

Example:

30cm3 of bleach was added to an excess of iodide ions, I- in sulphuric acid. 

Bleach contains chlorate (I) ions, ClO-.

In the presence of acid, chlorate (I) ions oxidise iodide ions, I- to iodine, I2

The iodine formed was titrated against 0.200 Mol dm-3 of sodium thiosulphate, 29.45cm3 was required.

Calculate the concentration of chlorate ions, ClO- in the bleach:

 

ClO-(aq) + 2H+(aq) + 2I-(aq)

Cl-(aq) + I2(laq) + H2O(aq) [1]
30.00cm3               BROWN  

Then:

2S2O32-(aq) + I2(aq)

2I-(aq) + S4O62-(aq)
    BROWN   COLOURLESS    
0.200 Mol dm-3            
29.45 cm3            
           
 
 

    

       
           
Moles = C x V/1000 2 : 1          
= 0.200 x 29.45/1000          
= 5.890 x 10-3 x 0.5

= 2.945 x 10-3

       

Estimating the copper content of solutions and alloys:

Practical

            1)  Iodine is liberated from it ions - using Cu2+ ions: 

2Cu2+(aq) + 4I-  2(aq)

2CuI(s) + I2(aq)
    COLOURLESS   WHITE SOLID   BROWN SOLn

            2)  The liberated iodine is titrated with a known concentration of thiosulphate,

 

2S2O32-(aq) + I2(aq)

2I-(aq) + S4O62-(aq)
    BROWN   COLOURLESS    

Copper metal:

1)  Convert Cu to Cu2+:

  conc. HNO3          
Cu(s)

Cu2+(aq) + 2e-   [1]
                 

2)  Liberate iodine:

2Cu2+(aq) + 4I-  2(aq)

2CuI(s) + I2(aq)
    COLOURLESS   WHITE SOLID   BROWN SOLn

           

3)  Titrate with a known concentration of thiosulphate,

 

2S2O32-(aq) + I2(aq)

2I-(aq) + S4O62-(aq)
    BROWN   COLOURLESS    

Ratio:                  S2O32-  :  I2  :  Cu2+  :  Cu

                                 2        :  1   :    2       :   2

Overall ratio is      1                 :             1

Example:

0.500g of bronze was reacted with nitric acid giving a Cu2+ solution. 

The solution was reacted with iodide ions, I- forming a solution of iodine, I2.

This iodine solution, I2 was then titrated with 0.200 mol dm-3 solution of sodium thiosulphate, S2O32-.  22.40cm3 of this was required.

Calculate the % copper in brass:

 

1)  Convert Cu in brass to Cu2+:

  conc. HNO3          
Cu(s)

Cu2+(aq) + 2e-   [1]

 

2)  Liberate iodine:

2Cu2+(aq) + 4I-  2(aq)

2CuI(s) + I2(aq)
    COLOURLESS   WHITE SOLID   BROWN SOLn

           

3)  Titrate with a known concentration of thiosulphate,

2S2O32-(aq) + I2(aq)

2I-(aq) + S4O62-(aq)
    BROWN   COLOURLESS    
0.200 Mol dm-3            
22.4 cm3            
           
 
 

    

       
           
Moles = C x V/1000            
= 0.200 x 22.40/1000            
= 4.48 x 10-3            

Qu 1 - 2  P 223 / Qu 9  P 225 / Qu's 226 - 229