`Mg^(+2)` does not form either preoxide or superoxidem because

Options

A. `Mg^(+2)` ion is relatively bigger
B. `Mg^(+2)` ion is relatively smaller
C. `Mg^(+2)` ion is stable
D. `Mg^(+2)` ion is unstable

Answer

To explain why \( \text{Mg}^{2+} \) does not form either peroxides or superoxides, we can follow these steps:

### Step 1: Understand the Electronic Configuration of Magnesium
Magnesium (Mg) has the atomic number 12. Its electronic configuration is:
\[ \text{Mg: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \]
When magnesium loses two electrons to form \( \text{Mg}^{2+} \), its electronic configuration becomes:
\[ \text{Mg}^{2+}: 1s^2 \, 2s^2 \, 2p^6 \quad \text{(or simply, } 2, 8 \text{)} \]

### Step 2: Analyze the Size of the \( \text{Mg}^{2+} \) Ion
The \( \text{Mg}^{2+} \) ion is relatively small due to the loss of its outermost electrons. The effective nuclear charge experienced by the remaining electrons is higher, leading to a stronger attraction between the nucleus and the electrons, thus reducing the size of the ion.

### Step 3: Understand the Formation of Peroxides and Superoxides
Peroxides (e.g., \( \text{O}_2^{2-} \)) and superoxides (e.g., \( \text{O}_2^{-} \)) are typically formed by larger cations. These larger cations can stabilize the larger anions (peroxide and superoxide) due to their ability to accommodate the increased electron density and the larger ionic size.

### Step 4: Relate Ion Size to Stability of Peroxides and Superoxides
Since \( \text{Mg}^{2+} \) is a small ion, it cannot stabilize the larger peroxide or superoxide ions. The small size of \( \text{Mg}^{2+} \) means that it does not have the necessary space or ability to effectively interact with the larger \( \text{O}_2^{2-} \) or \( \text{O}_2^{-} \) ions.

### Conclusion
Therefore, \( \text{Mg}^{2+} \) does not form either peroxides or superoxides because its small ionic size does not allow it to stabilize these larger anions.

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