Draw the molecular orbital diagram for No. Label all of the atomic orbitals and molecular orbitals and put the correct number of electrons in. You do not need to draw the shapes of any of the orbitals.
a) MO diagram
b) Based on your MO diagram, is N, diamagnetic or paramagnetic?
c) Calculate the bond order for N2.
The Correct Answer and Explanation is :
a) Drawing the MO Diagram for N₂:
For nitrogen (N₂), we need to consider the electronic configuration of nitrogen, which has 7 electrons in its outer shell. In the molecule N₂, we have a total of 14 electrons to distribute across the molecular orbitals.
- Atomic Orbitals:
Nitrogen has 2s and 2p orbitals, so we have: - 2 atomic 2s orbitals
- 4 atomic 2p orbitals (2pz, 2px, and 2py)
- Molecular Orbitals:
The molecular orbitals are formed by the combination of atomic orbitals from each nitrogen atom. These include: - σ(2s), σ*(2s) (bonding and anti-bonding from the 2s orbitals)
- σ(2pz), π(2px), π(2py), π(2px), π(2py), σ*(2pz) (bonding and anti-bonding from the 2p orbitals)
- The order of energy levels for N₂ (since it is a molecule with atomic number 7) is:
- σ(2s) < σ(2s) < σ(2pz) < π(2px) = π(2py) < π(2px) = π(2py) < σ(2pz)
After the energy levels are set, electrons are placed in these orbitals in pairs (2 electrons per orbital). For N₂, you fill the orbitals as follows (starting from the lowest energy orbital):
- 2 electrons in σ(2s)
- 2 electrons in σ*(2s)
- 2 electrons in σ(2pz)
- 4 electrons in π(2px) and π(2py) (2 electrons in each)
- 2 electrons in π(2px) and π(2py)
b) Is N₂ Diamagnetic or Paramagnetic?
N₂ is diamagnetic because it has no unpaired electrons in its molecular orbitals. All electrons are paired, meaning it will not be attracted to a magnetic field.
c) Bond Order Calculation for N₂
Bond order is calculated as:
[
\text{Bond order} = \frac{1}{2} \left( \text{Number of bonding electrons} – \text{Number of anti-bonding electrons} \right)
]
From the MO diagram:
- Number of bonding electrons = 2 (σ(2s)) + 2 (σ(2pz)) + 4 (π(2px) and π(2py)) = 8 bonding electrons
- Number of anti-bonding electrons = 2 (σ(2s)) + 2 (π(2px) and π*(2py)) = 4 anti-bonding electrons
Thus:
[
\text{Bond order} = \frac{1}{2} (8 – 4) = \frac{1}{2} \times 4 = 2
]
The bond order for N₂ is 2, meaning it has a double bond.
I’ll now generate the molecular orbital diagram image for you.
Here is the molecular orbital diagram for the nitrogen molecule (N₂), as requested. It shows the atomic orbitals, molecular orbitals, and the placement of electrons in the corresponding orbitals. The bond order is 2, confirming a double bond between the nitrogen atoms.
If you have any further questions or need more details, feel free to ask!
