The molecular orbital configuration of a diatomic molecule is
`sigma 1s^(2) sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2) sigma2p_(z)^(2)pi2p_(x)^(2)pi2p_(y)^(2)`
Its bond order is _________.

Options

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

Answer

To find the bond order of the diatomic molecule given its molecular orbital configuration, we will follow these steps:

### Step 1: Identify the molecular orbital configuration
The given molecular orbital configuration is:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y²

### Step 2: Determine the number of electrons in bonding and anti-bonding orbitals
- **Bonding orbitals**:
- σ1s: 2 electrons
- σ2s: 2 electrons
- σ2p_z: 2 electrons
- π2p_x: 2 electrons
- π2p_y: 2 electrons
- Total in bonding orbitals = 2 + 2 + 2 + 2 + 2 = 10 electrons

- **Anti-bonding orbitals**:
- σ*1s: 2 electrons
- σ*2s: 2 electrons
- Total in anti-bonding orbitals = 2 + 2 = 4 electrons

### Step 3: Apply the bond order formula
The bond order (B.O.) is calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} \left( \text{Number of electrons in bonding MOs} - \text{Number of electrons in anti-bonding MOs} \right) \]

Substituting the values we found:
\[ \text{Bond Order} = \frac{1}{2} \left( 10 - 4 \right) \]
\[ \text{Bond Order} = \frac{1}{2} \times 6 = 3 \]

### Conclusion
The bond order of the diatomic molecule is **3**.

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