Complete the molecular orbital diagram of F2 and F2 What is the bond order of F2 ?
The Correct Answer and Explanation is:
To complete the molecular orbital diagram for F₂ (Fluorine molecule), we need to consider the molecular orbitals formed by the atomic orbitals of two fluorine atoms. Fluorine has an atomic number of 9, so each atom has 9 electrons, and for F₂, we have a total of 18 electrons.
The molecular orbital diagram for diatomic molecules like F₂ can be derived from the combination of atomic orbitals (σ, σ*, π, and π* orbitals). Here’s the procedure for completing the molecular orbital diagram for F₂:
Step 1: Molecular Orbitals for F₂
The orbitals involved in bonding for F₂ are:
- σ(2s): bonding orbital formed by the overlap of the 2s orbitals from each atom.
- *σ(2s)**: anti-bonding orbital formed by the overlap of the 2s orbitals from each atom, but out of phase.
- σ(2p_z): bonding orbital formed by the overlap of the 2p orbitals along the internuclear axis.
- π(2p_x) and π(2p_y): degenerate bonding orbitals formed by the sideways overlap of the 2p orbitals in the x and y directions.
- π(2p_x) and π(2p_y)**: degenerate anti-bonding orbitals formed by the sideways overlap of the 2p orbitals in the x and y directions, but out of phase.
- *σ(2p_z)**: anti-bonding orbital formed by the overlap of the 2p orbitals along the internuclear axis, but out of phase.
Step 2: Electron Configuration for F₂
Now, distribute the 18 electrons across these molecular orbitals, starting from the lowest energy orbital:
- σ(2s) (2 electrons)
- σ(2s)* (2 electrons)
- π(2p_x) and π(2p_y) (4 electrons)
- σ(2p_z) (2 electrons)
- π(2p_x)* and π(2p_y)* (4 electrons)
- σ(2p_z)* (2 electrons)
Step 3: Bond Order Calculation
Bond order is calculated as: Bond Order=12×(Number of bonding electrons−Number of anti-bonding electrons)\text{Bond Order} = \frac{1}{2} \times (\text{Number of bonding electrons} – \text{Number of anti-bonding electrons})Bond Order=21×(Number of bonding electrons−Number of anti-bonding electrons)
For F₂:
- Bonding electrons: 2 (σ(2s)) + 4 (π(2p_x) + π(2p_y)) + 2 (σ(2p_z)) = 8
- Anti-bonding electrons: 2 (σ*(2s)) + 4 (π*(2p_x) + π*(2p_y)) + 2 (σ*(2p_z)) = 8
So, the bond order for F₂ is: Bond Order=12×(8−8)=0\text{Bond Order} = \frac{1}{2} \times (8 – 8) = 0Bond Order=21×(8−8)=0
Thus, F₂ has a bond order of 1, meaning it has a single bond between the two fluorine atoms.
Conclusion
The molecular orbital diagram for F₂ shows that the electrons are evenly distributed between bonding and anti-bonding orbitals, which results in a bond order of 1. This bond order implies a stable bond between the two fluorine atoms. The low bond order is a characteristic feature of molecules with relatively high electronegativity like fluorine.
