Correct B.O. of `O_(2)` is :

Options

A. 1
B. 2
C. `2.5`
D. 3

Answer

To determine the bond order (B.O.) of the O₂ molecule, we can follow these steps:

### Step 1: Determine the Number of Electrons
Oxygen (O) has an atomic number of 8, which means each oxygen atom has 8 electrons. Therefore, O₂, which consists of two oxygen atoms, has a total of:
\[ 8 \text{ electrons} \times 2 = 16 \text{ electrons} \]

### Step 2: Write the Molecular Orbital Configuration
The molecular orbital configuration for O₂ can be written as follows:
1. Fill the bonding orbitals first:
- \( \sigma_{1s}^2 \)
- \( \sigma^*_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^*_{2s}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
2. Now fill the anti-bonding orbitals:
- \( \pi^*_{2p_x}^1 \)
- \( \pi^*_{2p_y}^1 \)

So, the complete configuration 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 3: Count the Number of Bonding and Anti-bonding Electrons
- **Bonding Electrons**:
- From the configuration, the bonding electrons are:
- \( \sigma_{1s}^2 \) (2)
- \( \sigma_{2s}^2 \) (2)
- \( \sigma_{2p_z}^2 \) (2)
- \( \pi_{2p_x}^2 \) (2)
- \( \pi_{2p_y}^2 \) (2)

Total bonding electrons = \( 2 + 2 + 2 + 2 + 2 = 10 \)

- **Anti-bonding Electrons**:
- From the configuration, the anti-bonding electrons are:
- \( \sigma^*_{1s}^2 \) (2)
- \( \sigma^*_{2s}^2 \) (2)
- \( \pi^*_{2p_x}^1 \) (1)
- \( \pi^*_{2p_y}^1 \) (1)

Total anti-bonding electrons = \( 2 + 2 + 1 + 1 = 6 \)

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

### Conclusion
The bond order of O₂ is 2, indicating that there is a double bond between the two oxygen atoms.

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