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 Total Number of Electrons
Oxygen has an atomic number of 8, which means each oxygen atom has 8 electrons. Since \( O_2 \) consists of two oxygen atoms, the total number of electrons is:
\[ 8 \times 2 = 16 \text{ electrons} \]
However, since we have a negative charge (\( O_2^- \)), we add one more electron:
\[ 16 + 1 = 17 \text{ electrons} \]

### 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 \( O_2^- \), we need to add one more electron to the configuration. This additional electron will go into the next available molecular orbital, which is the \( \pi^*_{2p_x} \) or \( \pi^*_{2p_y} \) orbital. Thus, the configuration for \( O_2^- \) becomes:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^2 \]

### Step 3: Count the Bonding and Antibonding Electrons
- **Bonding Electrons**:
- \( \sigma_{1s}^2 \) = 2
- \( \sigma_{2s}^2 \) = 2
- \( \sigma_{2p_z}^2 \) = 2
- \( \pi_{2p_x}^2 \) = 2
- \( \pi_{2p_y}^2 \) = 2
- Total bonding electrons = \( 2 + 2 + 2 + 2 + 2 = 10 \)

- **Antibonding Electrons**:
- \( \sigma^*_{1s}^2 \) = 2
- \( \sigma^*_{2s}^2 \) = 2
- \( \pi^*_{2p_x}^2 \) = 2
- Total antibonding electrons = \( 2 + 2 + 2 = 6 \)

### Step 4: Calculate the Bond Order
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons})}{2} \]
Substituting the values we found:
\[ \text{Bond Order} = \frac{(10 - 6)}{2} = \frac{4}{2} = 2 \]

### Step 5: Conclusion
The bond order of \( O_2^- \) is 1.5.