The bond order of : `NO` is

Explanation

To determine the bond order of nitric oxide (NO), we can follow these steps:

### Step 1: Determine the total number of electrons in NO
Nitric oxide (NO) consists of nitrogen (N) and oxygen (O). The atomic number of nitrogen is 7, and for oxygen, it is 8. Therefore, the total number of electrons in NO is:
\[ 7 \text{ (from N)} + 8 \text{ (from O)} = 15 \text{ electrons} \]

### Step 2: Write the molecular orbital (MO) configuration for NO
The molecular orbital configuration for NO can be written as follows:
- Start with the lowest energy level:
- \( \sigma_{1s}^2 \)
- \( \sigma^*_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^*_{2s}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \pi^*_{2p_x}^1 \) (the remaining electron goes here)
- \( \pi^*_{2p_y}^0 \)

Thus, the complete MO configuration of NO is:
\[ \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1 \]

### Step 3: Count the number of bonding and anti-bonding electrons
- **Bonding Electrons**:
- \( \sigma_{1s}^2 \) = 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**:
- \( \sigma^*_{1s}^2 \) = 2
- \( \sigma^*_{2s}^2 \) = 2
- \( \pi^*_{2p_x}^1 \) = 1
- \( \pi^*_{2p_y}^0 \) = 0
- Total anti-bonding electrons = \( 2 + 2 + 1 = 5 \)

### Step 4: 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 in the values we found:
\[ \text{Bond Order} = \frac{10 - 5}{2} = \frac{5}{2} = 2.5 \]

### Conclusion
The bond order of nitric oxide (NO) is 2.5.

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