The bond order of `O_2^(-)` is:

Options

A. `0.5`
B. `1.5`
C. `3.5`
D. `2.5`

Explanation

To find the bond order of the ion \( O_2^{-} \), we will follow these steps:

### Step 1: Determine the number of electrons
The oxygen atom has an atomic number of 8, which means each oxygen atom has 8 electrons. Therefore, the diatomic oxygen molecule \( O_2 \) has:
\[ 8 \text{ (from first O)} + 8 \text{ (from second O)} = 16 \text{ electrons} \]
Since we are considering the \( O_2^{-} \) ion, we need to add one more electron:
\[ 16 + 1 = 17 \text{ electrons} \]

### Step 2: Write the molecular orbital configuration
Using molecular orbital theory, the electron configuration for \( O_2^{-} \) can be written as follows:
1. \( \sigma_{1s}^2 \)
2. \( \sigma^*_{1s}^2 \)
3. \( \sigma_{2s}^2 \)
4. \( \sigma^*_{2s}^2 \)
5. \( \sigma_{2p_z}^2 \)
6. \( \pi_{2p_x}^2 \)
7. \( \pi_{2p_y}^2 \)
8. \( \pi^*_{2p_x}^1 \) (the additional electron goes here)
9. \( \pi^*_{2p_y}^0 \)

This can be summarized as:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \]

### Step 3: Count the electrons in bonding and anti-bonding orbitals
- **Bonding Molecular Orbitals (BMOs)**:
- \( \sigma_{1s}^2 \) = 2
- \( \sigma_{2s}^2 \) = 2
- \( \sigma_{2p_z}^2 \) = 2
- \( \pi_{2p_x}^2 \) = 2
- \( \pi_{2p_y}^2 \) = 2
- **Total in BMOs** = \( 2 + 2 + 2 + 2 + 2 = 10 \)

- **Anti-bonding Molecular Orbitals (ABMOs)**:
- \( \sigma^*_{1s}^2 \) = 2
- \( \sigma^*_{2s}^2 \) = 2
- \( \pi^*_{2p_x}^1 \) = 1
- \( \pi^*_{2p_y}^0 \) = 0
- **Total in ABMOs** = \( 2 + 2 + 1 = 5 \)

### Step 4: Calculate the bond order
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} \left( \text{Number of electrons in BMOs} - \text{Number of electrons in ABMOs} \right) \]
Substituting the values we found:
\[ \text{Bond Order} = \frac{1}{2} (10 - 5) = \frac{1}{2} \times 5 = 2.5 \]

### Final Answer
Thus, the bond order of \( O_2^{-} \) is **1.5**.

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