`NO^(+)` has bond order

Options

A. 2
B. 2.5
C. 3
D. 3.5

Explanation

To determine the bond order of the NO⁺ ion, we will follow these steps:

### Step 1: Determine the total number of electrons in NO⁺
NO⁺ is a nitrogen-oxide ion that has lost one electron compared to neutral NO. The neutral NO molecule has a total of 15 electrons (7 from nitrogen and 8 from oxygen). Therefore, NO⁺ has:
\[ 15 - 1 = 14 \text{ electrons} \]

### Step 2: Write the molecular orbital configuration for NO⁺
The molecular orbital configuration for NO⁺, which has 14 electrons, is as follows:
- \(\sigma 1s^2\)
- \(\sigma^* 1s^2\)
- \(\sigma 2s^2\)
- \(\sigma^* 2s^2\)
- \(\pi 2p_x^2\)
- \(\pi 2p_y^2\)
- \(\sigma 2p_z^2\)

This configuration indicates the distribution of electrons in the molecular orbitals.

### Step 3: Identify bonding and anti-bonding molecular orbitals
In the configuration:
- **Bonding molecular orbitals**: \(\sigma 1s\), \(\sigma 2s\), \(\pi 2p_x\), \(\pi 2p_y\), \(\sigma 2p_z\)
- **Anti-bonding molecular orbitals**: \(\sigma^* 1s\), \(\sigma^* 2s\)

### Step 4: Count the number of electrons in bonding and anti-bonding orbitals
- **Electrons in bonding molecular orbitals**:
- \(\sigma 1s^2\) = 2
- \(\sigma 2s^2\) = 2
- \(\pi 2p_x^2\) = 2
- \(\pi 2p_y^2\) = 2
- \(\sigma 2p_z^2\) = 2
- Total = \(2 + 2 + 2 + 2 + 2 = 10\) electrons in bonding orbitals.

- **Electrons in anti-bonding molecular orbitals**:
- \(\sigma^* 1s^2\) = 2
- \(\sigma^* 2s^2\) = 2
- Total = \(2 + 2 = 4\) electrons in anti-bonding orbitals.

### Step 5: Apply the bond order formula
The bond order (BO) can be calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of electrons in bonding MOs}) - (\text{Number of electrons in anti-bonding MOs})}{2} \]
Substituting the values we found:
\[ \text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3 \]

### Conclusion
The bond order of NO⁺ is 3.

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