The oxidation states of `Cu` and `Fe` in chalcopyrite are, respectively,

Options

A. `+2,+2`
B. `+1,+2`
C. `+1,+3`
D. `+2,+1`

Explanation

To determine the oxidation states of copper (Cu) and iron (Fe) in chalcopyrite (CuFeS2), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula**: The chemical formula for chalcopyrite is CuFeS2.

2. **Assign Oxidation States to Sulfur**: In chalcopyrite, sulfur (S) typically has an oxidation state of -2. Since there are two sulfur atoms in the formula, the total contribution from sulfur is:
\[ 2 \times (-2) = -4 \]

3. **Set Up the Equation**: The sum of the oxidation states in a neutral compound must equal zero. Therefore, we can set up the following equation based on the oxidation states of Cu and Fe:
\[ \text{Oxidation state of Cu} + \text{Oxidation state of Fe} + \text{Total oxidation state from S} = 0 \]
Let the oxidation state of Cu be \( x \) and the oxidation state of Fe be \( y \). The equation becomes:
\[ x + y - 4 = 0 \]

4. **Consider Known Oxidation States**: We know that copper can have oxidation states of +1 or +2, and iron can have oxidation states of +2 or +3.

5. **Assume Oxidation States**: Let's assume the oxidation state of copper is +1 (x = +1). Then we can substitute this value into our equation:
\[ +1 + y - 4 = 0 \]
Simplifying this gives:
\[ y - 3 = 0 \implies y = +3 \]

6. **Conclusion**: Therefore, the oxidation states of copper and iron in chalcopyrite are:
- Copper (Cu): +1
- Iron (Fe): +3

### Final Answer:
The oxidation states of Cu and Fe in chalcopyrite are +1 and +3, respectively.

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