Name a molecule for which bond order is 3.

Answer

To find a molecule with a bond order of 3, we can analyze the nitrogen molecule (N₂) as an example. Here’s a step-by-step solution:

### Step 1: Understand Bond Order
Bond order is defined as the number of bonding pairs of electrons between two atoms. It can be calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})}{2} \]

### Step 2: Write the Electronic Configuration of N₂
The electronic configuration of the nitrogen molecule (N₂) can be derived from the atomic configurations of nitrogen atoms. Each nitrogen atom has 7 electrons, so for N₂, we have a total of 14 electrons.

The molecular orbital configuration for N₂ is:
- \( \sigma_{1s}^2 \)
- \( \sigma^*_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^*_{2s}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)

### Step 3: Count the Bonding and Anti-bonding Electrons
From the molecular orbital configuration:
- **Bonding Electrons**:
- \( \sigma_{1s}^2 = 2 \)
- \( \sigma_{2s}^2 = 2 \)
- \( \pi_{2p_x}^2 + \pi_{2p_y}^2 = 2 + 2 = 4 \)
- \( \sigma_{2p_z}^2 = 2 \)

Total bonding electrons = \( 2 + 2 + 4 + 2 = 10 \)

- **Anti-bonding Electrons**:
- \( \sigma^*_{1s}^2 = 2 \)
- \( \sigma^*_{2s}^2 = 2 \)

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

### Step 4: Calculate the Bond Order
Now, we can substitute the values into the bond order formula:
\[ \text{Bond Order} = \frac{(10 - 4)}{2} = \frac{6}{2} = 3 \]

### Conclusion
Thus, the bond order of the nitrogen molecule (N₂) is 3.

### Final Answer
The molecule for which the bond order is 3 is **N₂ (Nitrogen molecule)**.

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