Problem on Oxidation state

Question:

What are the oxidation state of the metal Fe in the minerals haematite and magnetite?

Options:

a)      II, III in haematite and II in magnetite

b)      II, III in haematite and III in magnetite

c)      II in haematite and II, II in magnetite

d)     III in haematite and II, III in magnetite

Solution:

Finding the oxidation state of a metal in the mineral is easy if we know the chemical formula of mineral.

The chemical formula of haematite is Fe₂O₃.

Let us assume oxidation state of Fe is x.

The oxidation state of oxygen is -2.

Thus oxidation state of Fe in haematite: 2(x) + 3 (-2) = 0

                                                                2x – 6 = 0

                                                                  x = 3

The oxidation state of Fe in haematite is +3.

Now let us find out oxidation state of Fe in magnetite i.e. in Fe₃O₄

                                                       3(x) + 4 (-2) = 0  

                                                       3x – 8 = 0

                                                        X = 8 / 3

Thus according to calculation, the oxidation state of Fe in Fe₃O₄ is 8/3. But no single Fe has a charge of +8/3 because magnetite is an equimolar mixture of FeO and Fe₂O₃ which are combined in a certain way.

The oxidation state of Fe in FeO is +2

The oxidation state of Fe₂O₃ is +3 (as calculated earlier)

The formal oxidation state of Fe₃O₄ to be 8/3 is due to relative amount of Fe2+ of FeO and Fe3+ of Fe₂O₃. So, oxidation state of Fe in magnetite is +2 and +3.

 Thus option D is correct.