The ground state eletronic configuration of N2 molecule is written as: `KK(sigma _(2s))^2(sigma_(2s)^(**))^2(pi_(2px))^2=(pi_(2py))^2(sigma_(2pz))^2` The bond order of `N_2` is

Options

A. 3
B. 2
C. 0
D. 1

Answer

To determine the bond order of the N2 molecule from its electronic configuration, we can follow these steps:

### Step 1: Identify the Electronic Configuration
The given electronic configuration for the N2 molecule is:
\[ \text{KK}(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2px})^2=(\pi_{2py})^2(\sigma_{2pz})^2 \]

### Step 2: Count the Electrons in Bonding and Anti-bonding Orbitals
From the configuration, we can identify the following:

- **Bonding Molecular Orbitals (BMOs)**:
- \((\sigma_{2s})^2\) contributes 2 electrons
- \((\pi_{2px})^2\) contributes 2 electrons
- \((\pi_{2py})^2\) contributes 2 electrons
- \((\sigma_{2pz})^2\) contributes 2 electrons

Total electrons in bonding orbitals = \(2 + 2 + 2 + 2 = 8\)

- **Anti-bonding Molecular Orbitals (ABMOs)**:
- \((\sigma^*_{2s})^2\) contributes 2 electrons

Total electrons in anti-bonding orbitals = \(2\)

### Step 3: Calculate the Bond Order
The bond order is 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:
\[ \text{Bond Order} = \frac{8 - 2}{2} = \frac{6}{2} = 3 \]

### Conclusion
The bond order of the N2 molecule is **3**.

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