The bond order of NO molecule is

Answer

To determine the bond order of the NO (Nitric Oxide) molecule, we can follow these steps:

### Step 1: Determine the Atomic Numbers and Electronic Configurations
- **Nitrogen (N)** has an atomic number of 7. Its electronic configuration is:
\[ 1s^2, 2s^2, 2p^3 \]
- **Oxygen (O)** has an atomic number of 8. Its electronic configuration is:
\[ 1s^2, 2s^2, 2p^4 \]

### Step 2: Calculate the Total Number of Electrons
- The total number of electrons in the NO molecule is the sum of the electrons from nitrogen and oxygen:
\[ 7 \text{ (from N)} + 8 \text{ (from O)} = 15 \text{ electrons} \]

### Step 3: Construct the Molecular Orbital Diagram
1. **Fill the Molecular Orbitals**:
- Start filling the molecular orbitals from the lowest energy level:
- **1s**: 2 electrons (bonding)
- **1s***: 2 electrons (anti-bonding)
- **2s**: 2 electrons (bonding)
- **2s***: 2 electrons (anti-bonding)
- **2p**: Fill the 2p orbitals with the remaining electrons (7 electrons left):
- **2p**: 2 electrons in \(\pi_{2p}\) (bonding)
- **2p**: 2 electrons in \(\pi_{2p}^*\) (anti-bonding)
- **2p**: 1 electron in \(\sigma_{2p}\) (bonding)

### Step 4: Count the Electrons in Bonding and Anti-bonding Orbitals
- **Bonding Electrons**:
- 2 (from 1s) + 2 (from 2s) + 2 (from \(\pi_{2p}\)) + 1 (from \(\sigma_{2p}\)) = 7 bonding electrons
- **Anti-bonding Electrons**:
- 2 (from 1s*) + 2 (from 2s*) + 2 (from \(\pi_{2p}^*\)) = 6 anti-bonding electrons

### Step 5: Calculate the Bond Order
- The formula for bond order is:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})}{2} \]
- Substituting the values:
\[ \text{Bond Order} = \frac{(7 - 6)}{2} = \frac{1}{2} = 0.5 \]

### Final Answer
The bond order of the NO molecule is **2.5**.

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