Oxidation states of the metal in the minerals haematite and magnetite , respectively , are

Options

A. II,III in haematite and III in magnetite
B. II,III in haematite and II in magnetite
C. II in haematite and II, III in magnetite
D. III in haematite and II, III in magnetite

Answer

To determine the oxidation states of the metal in the minerals haematite and magnetite, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Chemical Formulas:**
- Haematite is represented by the formula \( \text{Fe}_2\text{O}_3 \).
- Magnetite is represented by the formula \( \text{Fe}_3\text{O}_4 \).

2. **Calculate the Oxidation State in Haematite:**
- For haematite (\( \text{Fe}_2\text{O}_3 \)):
- Let the oxidation state of iron (Fe) be \( x \).
- The oxidation state of oxygen (O) is \(-2\).
- The equation for the compound is:
\[ 2x + 3(-2) = 0 \]
- Simplifying this gives:
\[ 2x - 6 = 0 \implies 2x = 6 \implies x = +3 \]
- Thus, the oxidation state of iron in haematite is \( +3 \).

3. **Calculate the Oxidation State in Magnetite:**
- For magnetite (\( \text{Fe}_3\text{O}_4 \)):
- The formula can be viewed as \( \text{FeO} \cdot \text{Fe}_2\text{O}_3 \).
- Let the oxidation states of the three iron atoms be \( x_1, x_2, x_3 \).
- In magnetite, one iron is in the \( +2 \) state and two are in the \( +3 \) state.
- Therefore, we can set up the equation:
\[ x_1 + 2x_2 + 4(-2) = 0 \]
- Substituting \( x_1 = +2 \) and \( x_2 = +3 \):
\[ +2 + 2(+3) - 8 = 0 \implies +2 + 6 - 8 = 0 \]
- Thus, the oxidation states of iron in magnetite are \( +2 \) and \( +3 \).

4. **Conclusion:**
- The oxidation state of iron in haematite is \( +3 \).
- The oxidation states of iron in magnetite are \( +2 \) and \( +3 \).

### Final Answer:
The oxidation states of the metal in haematite and magnetite are \( +3 \) and \( +2, +3 \) respectively.