According to `MOT` whch of the following statement about magnetic character and bond order is corrent regarding `O_(2)^(o+)` .

Options

A. Paramagnetic and bond order `ltO_(2).`
B. Paramagnetic and bond order `gtO_(2).`
C. Diamagnetic and bond order `ltO_(2)`.
D. Diamagnetic and bond order `gtO_(2).`

Answer

To determine the magnetic character and bond order of \( O_2^+ \) according to Molecular Orbital Theory (MOT), we will follow these steps:

### Step 1: Determine the Total Number of Electrons
- For \( O_2 \), the total number of electrons is 16 (8 from each oxygen atom).
- For \( O_2^+ \), one electron is removed, resulting in a total of 15 electrons.
- For \( O_2^- \), one electron is added, resulting in a total of 17 electrons.

### Step 2: Write the Molecular Orbital Configuration
- The molecular orbital configuration for \( O_2 \) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \]
- For \( O_2^+ \) (15 electrons):
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^0 \]
- For \( O_2^- \) (17 electrons):
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^1 \]

### Step 3: Count Bonding and Antibonding Electrons
- For \( O_2^+ \):
- Bonding electrons: \( 10 \) (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} \))
- Antibonding electrons: \( 5 \) (from \( \pi_{2p_x}^* \))

- For \( O_2^- \):
- Bonding electrons: \( 10 \)
- Antibonding electrons: \( 7 \) (from \( \pi_{2p_x}^* \) and \( \pi_{2p_y}^* \))

### Step 4: Calculate Bond Order
- Bond order formula:
\[ \text{Bond Order} = \frac{\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons}}{2} \]
- For \( O_2^+ \):
\[ \text{Bond Order} = \frac{10 - 5}{2} = 2.5 \]
- For \( O_2^- \):
\[ \text{Bond Order} = \frac{10 - 7}{2} = 1.5 \]

### Step 5: Determine Magnetic Character
- \( O_2^+ \) has an unpaired electron in the antibonding orbital (\( \pi_{2p_x}^* \)), making it **paramagnetic**.
- \( O_2^- \) has all electrons paired, making it **diamagnetic**.

### Conclusion
- The correct statement regarding \( O_2^+ \) is that it is **paramagnetic** and has a **bond order greater than \( O_2^- \)**.

### Final Answer
Thus, the correct option is: **paramagnetic and bond order greater than \( O_2^- \)**.

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