Using MOT, compare `O_2^(+) and O_2^(-)` species and choose the incorrect option

Options

A. `O_2^(+)` is diamagnetic while `O_2^(-)` is paramagnetic
B. `O_2^(+)` have higher bond order than `O_2^(-)`
C. Both `O_2^(+) and O_2^(-)` are paramagnetic
D. `O_2^(-)` is less stable

Explanation

To compare the species \( O_2^{+} \) and \( O_2^{-} \) using Molecular Orbital Theory (MOT), we will analyze their electronic configurations, bond orders, magnetic behaviors, and relative stabilities.

### Step 1: Determine the Electronic Configuration of \( O_2 \)
The electronic configuration of \( O_2 \) (dioxygen) using MOT is:
\[ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*1} \pi_{2p_y}^{*1} \]

### Step 2: Determine the Electronic Configuration of \( O_2^{+} \)
For \( O_2^{+} \), we remove one electron from the highest energy molecular orbital, which is one of the \( \pi^* \) orbitals:
\[ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*1} \]

### Step 3: Calculate the Bond Order of \( O_2^{+} \)
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons}) - (\text{Number of antibonding electrons})}{2} \]
- Bonding electrons: 10 (from \( \sigma \) and \( \pi \) orbitals)
- Antibonding electrons: 5 (from \( \sigma^* \) and \( \pi^* \) orbitals)

Thus, the bond order for \( O_2^{+} \) is:
\[ \text{Bond Order} = \frac{10 - 5}{2} = 2.5 \]

### Step 4: Determine the Magnetic Behavior of \( O_2^{+} \)
Since \( O_2^{+} \) has one unpaired electron in the \( \pi^* \) orbital, it is **paramagnetic**.

### Step 5: Determine the Electronic Configuration of \( O_2^{-} \)
For \( O_2^{-} \), we add one electron to the highest energy molecular orbital, which is one of the \( \pi^* \) orbitals:
\[ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*2} \]

### Step 6: Calculate the Bond Order of \( O_2^{-} \)
Using the same bond order formula:
- Bonding electrons: 10
- Antibonding electrons: 6 (from \( \sigma^* \) and \( \pi^* \) orbitals)

Thus, the bond order for \( O_2^{-} \) is:
\[ \text{Bond Order} = \frac{10 - 6}{2} = 2 \]

### Step 7: Determine the Magnetic Behavior of \( O_2^{-} \)
Since \( O_2^{-} \) has no unpaired electrons, it is **diamagnetic**.

### Step 8: Compare Stability
The stability of these species can be compared based on their bond orders:
- \( O_2^{+} \) has a bond order of 2.5, while \( O_2^{-} \) has a bond order of 2.
- Higher bond order indicates greater stability.

### Conclusion
1. \( O_2^{+} \) is more stable than \( O_2^{-} \).
2. \( O_2^{+} \) is paramagnetic, while \( O_2^{-} \) is diamagnetic.
3. The incorrect option among the statements provided is that \( O_2^{-} \) is more stable than \( O_2^{+} \).