 Explanation of variable oxidation states:  All form +2 OS except Sc (loss of 4s electrons)  Max OS in theory is loss/use of 4s and 3d electrons.  This is achievable up to Mn (+7)  OS’s between +2 and max  Successive ionisation energies are quite similar

 Rules for working out Oxidation Numbers.  Uncombined elements: have oxidation numbers of  zero (0).  Simple Ions e.g. Na+:  the charge on the ion(1+) is the oxidation number of the element(+1)  so group 1, 2 and 3 elements have oxidation numbers of  1, 2 and 3 respectively when combined in compounds. 

 Fluorine: when combined ALWAYS has an oxidation number of  –1 (it is the most electronegative element).  Oxygen: when combined, has an oxidation number of  –2 EXCEPT  in peroxides (O 2 2- ) when it is –1 and when combined with fluorine when it will be positive e.g. OF 2 oxidation number is  +2.

 Hydrogen: when combined, usually has an oxidation number of  +1 unless  it is combined with a more electropositive metal, e.g. in metal hydrides such as sodium hydride NaH when it is  –1.  Chlorine: when combined, is always  –1 EXCEPT  when combined with OXYGEN, FLUORINE or NITROGEN.

In a compound: sum of OS’s = 0 For an ion: sum of OS’s = Charge on ion

 Li 2 S  Na 3 PO 4  K 2 Cr 2 O 7  Na 2 S 2 O 3  KFeCl 4  Na 4 [Fe(CN) 6 ]  Write the formula of the individual/free ions in each case.  Identify ligands where appropriate

 Change in oxidation state corresponds to  Number of electrons transferred  OIL RIG – oxidation corresponds to what type of change in OS?  More +ve/less –ve

 Fe 3+ changes to Fe 2+  OS change from +3  +2  Reduction, one electron involved  Fe 3+ + e -  Fe 2+ (half equation)  Sn 2+ changes to Sn 4+  Sn 2+  Sn e -

 NUMBER OF ELECTRONS LOST AND GAINED MUST BE THE SAME  So one or both half equations must be multiplied to satisfy this  Sn 2+  Sn e - : 2 e - involved  hence  2Fe e -  2Fe 2+

MnO 4 - reacting with Fe 2+ MnO 4 - is reduced to Mn 2+ Fe 2+ is oxidised to Fe 3+

 Use oxidation states to determine how many electrons are lost/gained  MnO 4 - Mn OS=  +7  Mn 2+ OS =  +2  MnO 4 - is reduced to Mn 2+ : 5e - s gained

 Fe 2+ is oxidised to Fe 3+  1 e - lost  5Fe 2+  Fe e -  i.e. 5 e - s lost

 5Fe 2+  Fe e  5e - s + MnO 4 -  Mn 2+  Reacting ratio 1 MnO 4 - : 5Fe 2+  Need to balance the oxygens:  All extra O’s become water  H’s required to do this are provided as H +

 5e - s + MnO 4 -  Mn H 2 O  8H + + 5e - s + MnO 4 -  Mn H 2 O  Combine this with:  5Fe 2+  Fe e -  8H + + 5Fe 2+ +MnO 4 -  Mn H 2 O +5Fe 3+