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 (O2

2-) when it is –1 and when combined with fluorine when it will be positive e.g. OF2 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 = 0For an ion: sum of OS’s = Charge on ion

Li2S Na3PO4

K2Cr2O7 Na2S2O3

KFeCl4 Na4[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

Fe3+ changes to Fe2+

OS change from +3 +2 Reduction, one electron involved Fe3+ + e- Fe2+ (half equation) Sn2+ changes to Sn4+

Sn2+ Sn4+ + 2e-

NUMBER OF ELECTRONS LOST AND GAINED MUST BE THE SAME

So one or both half equations must be multiplied to satisfy this

Sn2+ Sn4+ + 2e- : 2 e- involved hence 2Fe3+ + 2e- 2Fe2+

MnO4- reacting with Fe2+

MnO4- is reduced to Mn2+

Fe2+ is oxidised to Fe3+

Use oxidation states to determine how many electrons are lost/gained

MnO4- Mn OS=

+7 Mn2+ OS = +2 MnO4

- is reduced to Mn2+ : 5e-s gained

Fe2+ is oxidised to Fe3+

1 e- lost 5Fe2+ Fe3+ + 5e-

i.e. 5 e-s lost

5Fe2+ Fe3+ + 5e 5e-s + MnO4

- Mn2+

Reacting ratio 1 MnO4- : 5Fe2+

Need to balance the oxygens: All extra O’s become water H’s required to do this are provided as H+

5e-s + MnO4- Mn2+ + 4H2O

8H+ + 5e-s + MnO4- Mn2+ + 4H2O

Combine this with: 5Fe2+ Fe3+ + 5e-

8H+ + 5Fe2++MnO4-Mn2+ + 4H2O +5Fe3+