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 find the bond order of the N₂ molecule, we will follow these steps:

### Step 1: Understand the Concept of Bond Order
Bond order is defined as half the difference between the number of bonding electrons (nB) and the number of anti-bonding electrons (nA). The formula for bond order is given by:

\[ \text{Bond Order} = \frac{1}{2} (n_B - n_A) \]

### Step 2: Identify the Total Number of Electrons in N₂
The nitrogen atom has 7 electrons. Therefore, the N₂ molecule, which consists of two nitrogen atoms, has:

\[ 7 \text{ (from N)} + 7 \text{ (from N)} = 14 \text{ electrons} \]

### Step 3: Write the Ground State Electronic Configuration
The ground state electronic configuration of the N₂ molecule is given as:

\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \]

### Step 4: Count the Number of Bonding and Anti-bonding Electrons
From the electronic configuration:
- **Bonding Electrons (nB)**:
- \(\sigma_{1s}^2\): 2 electrons
- \(\sigma_{2s}^2\): 2 electrons
- \(\pi_{2p_x}^2\): 2 electrons
- \(\pi_{2p_y}^2\): 2 electrons
- \(\sigma_{2p_z}^2\): 2 electrons

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

- **Anti-bonding Electrons (nA)**:
- \(\sigma_{1s}^*^2\): 2 electrons
- \(\sigma_{2s}^*^2\): 2 electrons

Total anti-bonding electrons = \(2 + 2 = 4\)

### Step 5: Calculate the Bond Order
Now, substituting the values into the bond order formula:

\[ \text{Bond Order} = \frac{1}{2} (n_B - n_A) = \frac{1}{2} (10 - 4) = \frac{1}{2} \times 6 = 3 \]

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

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