To determine the oxidation states of iron in haematite (Fe₂O₃) and magnetite (Fe₃O₄), we need to analyze their chemical formulas and apply the concept of oxidation numbers.

Step 1: Understanding Oxidation State

The oxidation state (or oxidation number) of an element in a compound is the charge it would have if all bonds were ionic. For oxygen, the oxidation state is usually -2 (except in peroxides or superoxides).

Step 2: Oxidation State in Haematite (Fe₂O₃)

Let the oxidation state of Fe be $x$. Since oxygen has -2 oxidation state, the total charge from oxygen is $3\times (-2)=-6$. The compound is neutral, so the sum of oxidation states must be zero:

$2x+(-6)=0$

Solving: $2x=6$, so $x=3$. Thus, iron in haematite has an oxidation state of +3.

Step 3: Oxidation State in Magnetite (Fe₃O₄)

Magnetite is a mixed oxide containing both Fe²⁺ and Fe³⁺ ions. Its formula can be written as FeO·Fe₂O₃. Let the oxidation states be $y$ for some Fe and $z$ for others, but it's easier to calculate the average. For a neutral compound:

$3x+4\times (-2)=0$

So, $3x-8=0$, thus $3x=8$, and $x=\frac{8}{3}$. This fractional value indicates the presence of iron in two oxidation states: +2 and +3. Specifically, magnetite contains one Fe²⁺ and two Fe³⁺ ions.

Final Answer

Therefore, in haematite, iron has an oxidation state of III, and in magnetite, iron has oxidation states of II and III.

Comparing with the options: "III in haematite and II, III in magnetite" is correct.

Related Topics and Formulae

Oxidation Number Calculation: For a compound, the sum of oxidation numbers of all atoms equals the overall charge (zero for neutral compounds). Oxygen is usually -2.

Mixed Oxides: Some oxides, like magnetite, contain the same metal in multiple oxidation states.

Key Minerals: Haematite (Fe₂O₃) and magnetite (Fe₃O₄) are important ores of iron.