Using molecular orbital theory, calculate the bond order, molecular configuration, number of bonding electrons and number of antibonding electrons for CO and NO+

Answer 1

To analyze the molecular orbitals of CO (carbon monoxide) and NO\(^+\) (nitric oxide cation), we can employ Molecular Orbital (MO) theory. Let's break it down for each molecule one by one.

1. Carbon Monoxide (CO)

Atomic Numbers:

• Carbon (C) = 6 electrons (1s² 2s² 2p²)
• Oxygen (O) = 8 electrons (1s² 2s² 2p⁴)

Total Electrons in CO:

• Total = 6 (C) + 8 (O) = 14 electrons

Molecular Orbital Configuration:

For diatomic molecules like CO, the relevant molecular orbitals based on energy (for 2nd period elements) are organized as:

• σ(1s), σ*(1s), σ(2s), σ*(2s), σ(2p_z), π(2p_x), π(2p_y), π*(2p_x), π*(2p_y)

Filling in the molecular orbitals according to the Aufbau principle (from the lowest to the highest energy level):

1. σ(1s): 2 electrons
2. σ*(1s): 0 electrons
3. σ(2s): 2 electrons
4. σ*(2s): 0 electrons
5. σ(2p_z): 2 electrons
6. π(2p_x): 2 electrons
7. π(2p_y): 2 electrons
8. π*(2p_x): 0 electrons
9. π*(2p_y): 0 electrons

Filling:

So, the filling looks like this:

• σ(1s)² σ*(1s)⁰ σ(2s)² σ*(2s)⁰ σ(2p_z)² π(2p_x)² π(2p_y)²

Count Electrons:

• Bonding Electrons: 2 (σ(1s)) + 2 (σ(2s)) + 2 (σ(2p_z)) + 2 (π(2p_x)) + 2 (π(2p_y)) = 10 bonding electrons
• Antibonding Electrons: 0 (σ*(1s)) + 0 (σ*(2s)) + 0 (π*(2p_x)) + 0 (π*(2p_y)) = 0 antibonding electrons

Bond Order Calculation:

\[ \text{Bond Order} = \frac{N_b - N_a}{2} \] Where \(N_b\) = number of bonding electrons, \(N_a\) = number of antibonding electrons.

\[ \text{Bond Order for CO} = \frac{10 - 0}{2} = 5 \]

2. Nitric Oxide Cation (NO\(^+\))

Atomic Numbers:

• Nitrogen (N) = 7 electrons (1s² 2s² 2p³)
• Oxygen (O) = 8 electrons (1s² 2s² 2p⁴)

Total Electrons in NO\(^+\):

• Total = 7 (N) + 8 (O) - 1 (due to +1 charge) = 14 electrons

Molecular Orbital Configuration:

Similarly to CO, the molecular orbitals for NO are filled:

1. σ(1s): 2 electrons
2. σ*(1s): 0 electrons
3. σ(2s): 2 electrons
4. σ*(2s): 0 electrons
5. σ(2p_z): 2 electrons
6. π(2p_x): 2 electrons
7. π(2p_y): 2 electrons
8. π*(2p_x): 0 electrons
9. π*(2p_y): 0 electrons

Filling:

So, the filling looks like this:

• σ(1s)² σ*(1s)⁰ σ(2s)² σ*(2s)⁰ σ(2p_z)² π(2p_x)² π(2p_y)²

Count Electrons:

• Bonding Electrons: 2 (σ(1s)) + 2 (σ(2s)) + 2 (σ(2p_z)) + 2 (π(2p_x)) + 2 (π(2p_y)) = 10 bonding electrons
• Antibonding Electrons: 0 (σ*(1s)) + 0 (σ*(2s)) + 0 (π*(2p_x)) + 0 (π*(2p_y)) = 0 antibonding electrons

Bond Order Calculation:

\[ \text{Bond Order for NO}^+ = \frac{10 - 0}{2} = 5 \]

Summary of Results

• For CO:

  • Bond Order: 3
  • Bonding Electrons: 10
  • Antibonding Electrons: 0

• For NO\(^+\):

  • Bond Order: 3
  • Bonding Electrons: 10
  • Antibonding Electrons: 0

Both CO and NO\(^+\) have a bond order of 3 and share similar distributions of bonding and antibonding electrons.