The general formula of ferrites is `MFe_(2)O_(4)`. Where 'M' would not be

Options

A. `Mg`
B. `Cu`
C. `Al`
D. `Zn`

Answer

To solve the question regarding the general formula of ferrites, which is given as \( MFe_2O_4 \), we need to determine which metal 'M' cannot be present based on its oxidation state.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The formula \( MFe_2O_4 \) indicates that for every one metal ion 'M', there are two iron ions and four oxygen ions in the structure of ferrites.

2. **Oxidation States**:
In ferrites, iron (Fe) is typically in the +3 oxidation state. Therefore, we need to analyze the oxidation state of 'M' to maintain charge balance in the formula.

3. **Setting Up the Charge Balance**:
The total charge contributed by the iron ions in the formula is:
\[ 2 \times (+3) = +6 \]
The total charge contributed by the oxygen ions is:
\[ 4 \times (-2) = -8 \]
To find the oxidation state of 'M', we set up the equation for charge balance:
\[ X + 6 - 8 = 0 \]
Simplifying this gives:
\[ X - 2 = 0 \implies X = +2 \]
Thus, the oxidation state of 'M' must be +2.

4. **Identifying Possible Metals**:
Now we need to identify which metals can exist in the +2 oxidation state. Common metals that can have a +2 oxidation state include:
- Magnesium (Mg)
- Copper (Cu)
- Zinc (Zn)

5. **Identifying the Excluded Metal**:
We need to find a metal that does not exhibit a +2 oxidation state. Aluminum (Al), for example, typically has an oxidation state of +3. Therefore, aluminum cannot be 'M' in the formula \( MFe_2O_4 \).

6. **Conclusion**:
Based on the analysis, the metal 'M' in the formula \( MFe_2O_4 \) cannot be aluminum because it does not have a +2 oxidation state.

### Final Answer:
The metal 'M' cannot be aluminum.

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