Question

Calculate the bond order, the molecular configuration, Number of bonding electrons and Number of anti-bondig electrons for the following molecules.

1. Carbon monoxide?

2.Nitrosonium ?

Answer 1

To analyze carbon monoxide (CO) and nitrosonium ion (NO⁺), we can determine their molecular configurations, bond order, number of bonding electrons, and number of anti-bonding electrons using molecular orbital theory.

### 1. Carbon Monoxide (CO)

**Molecular Configuration:**
The molecular orbital configuration for \( CO \) can be derived from its atomic orbitals:
- Carbon (C) has the electronic configuration: \( 1s^2 2s^2 2p^2 \)
- Oxygen (O) has the electronic configuration: \( 1s^2 2s^2 2p^4 \)

Combining the atomic orbitals, we get the following filling in of molecular orbitals:
- \( \sigma_{1s}^2 \sigma_{1s}^2 \)
- \( \sigma_{2s}^2 \sigma_{2s}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \pi_{2p_y}^2 \) (both occupying the π orbitals)
- \( \pi_{2p_x}^0 \pi_{2p_y}^0 \)

Thus, the configuration of CO is:
\[ ( \sigma_{1s}^2) ( \sigma_{1s}^2) ( \sigma_{2s}^2) ( \sigma_{2s}^2) ( \sigma_{2p_z}^2) ( \pi_{2p_x}^2) ( \pi_{2p_y}^2) \]

**Bond Order Calculation:**
Bond Order = \( \frac{(Number \, of \, bonding \, electrons - Number \, of \, anti-bonding \, electrons)}{2} \)

- Bonding Electrons = 10 (2 from \( \sigma_{1s} \), 2 from \( \sigma_{2s} \), 2 from \( \sigma_{2p_z} \), 4 from \( \pi_{2p} \)).
- Anti-bonding Electrons = 0.

Bond Order = \( \frac{10 - 0}{2} = 5\)

**Number of Bonding and Anti-bonding Electrons:**
- Number of bonding electrons = 10
- Number of anti-bonding electrons = 0

### 2. Nitrosonium Ion (NO⁺)

**Molecular Configuration:**
The molecular orbital configuration for \( NO^+ \):
- Nitrogen (N) has the electronic configuration: \( 1s^2 2s^2 2p^3 \)
- Oxygen (O) has the electronic configuration: \( 1s^2 2s^2 2p^4 \)

However, since \( NO \) has 11 total valence electrons, removing one electron for the \( NO^+ \) ion leaves us with 10 valence electrons. The filling of molecular orbitals would be:
- \( \sigma_{1s}^2 \sigma_{1s}^2 \)
- \( \sigma_{2s}^2 \sigma_{2s}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \pi_{2p_y}^2 \)

Thus, the configuration for \( NO^+ \) is:
\[ ( \sigma_{1s}^2) ( \sigma_{1s}^2) ( \sigma_{2s}^2) ( \sigma_{2s}^2) ( \sigma_{2p_z}^2) ( \pi_{2p_x}^2) ( \pi_{2p_y}^2) \]

**Bond Order Calculation:**
Bond Order = \( \frac{(Number \, of \, bonding \, electrons - Number \, of \, anti-bonding \, electrons)}{2} \)

- Bonding Electrons = 10 (from filled orbitals)
- Anti-bonding Electrons = 0.

Bond Order = \( \frac{10 - 0}{2} = 5\)

**Number of Bonding and Anti-bonding Electrons:**
- Number of bonding electrons = 10
- Number of anti-bonding electrons = 0

### Summary
1. **Carbon Monoxide (CO)**
- Bond Order: 3
- Molecular Configuration: \( \sigma_{1s}^2 \sigma_{1s}^2 \sigma_{2s}^2 \sigma_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- Number of Bonding Electrons: 10
- Number of Anti-bonding Electrons: 0

2. **Nitrosonium Ion (NO⁺)**
- Bond Order: 3
- Molecular Configuration: \( \sigma_{1s}^2 \sigma_{1s}^2 \sigma_{2s}^2 \sigma_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- Number of Bonding Electrons: 10
- Number of Anti-bonding Electrons: 0