Draw a molecular orbital diagram for F2^2- . Calculate the bond order and magnetic behavior.
The Correct Answer and Explanation is :
Molecular Orbital (MO) Diagram for (F_2^{2-})
- Atomic Orbitals Combination:
- Fluorine has an atomic number of 9, so its electron configuration is (1s^2 2s^2 2p^5).
- In the molecular orbital diagram for diatomic fluorine, atomic orbitals combine to form molecular orbitals in the following order: [
\sigma(2s) < \sigma^(2s) < \pi(2p) < \sigma(2p) < \pi^(2p) < \sigma^*(2p)
] - This order follows from the energy interactions where, for elements with atomic numbers greater than 7, the (\sigma(2p)) orbital is lower in energy than the (\pi(2p)) orbitals.
- Electron Filling in (F_2^{2-}):
- The neutral (F_2) molecule has 18 valence electrons ((2 \times 9) from fluorine atoms, minus 2 from the filled (1s) core orbitals).
- The (2-) charge adds two more electrons, making a total of 20 valence electrons.
- These electrons are distributed among the molecular orbitals as follows:
- ( \sigma(2s) ): 2 electrons
- ( \sigma^*(2s) ): 2 electrons
- ( \pi(2p) ): 4 electrons
- ( \sigma(2p) ): 2 electrons
- ( \pi^*(2p) ): 4 electrons
- ( \sigma^*(2p) ): 2 electrons
Bond Order Calculation
Bond order is given by:
[
\text{Bond Order} = \frac{\text{Bonding electrons} – \text{Antibonding electrons}}{2}
]
- Bonding electrons: (2+4+2 = 8)
- Antibonding electrons: (2+4+2 = 8)
[
\text{Bond Order} = \frac{8 – 8}{2} = 0
]
Since the bond order is zero, the (F_2^{2-}) ion does not exist as a stable molecule.
Magnetic Behavior
A molecule is paramagnetic if it has unpaired electrons and diamagnetic if all electrons are paired.
- In (F_2^{2-}), all electrons are paired.
- Thus, (F_2^{2-}) is diamagnetic.
Conclusion
- MO Diagram: Shows complete filling of bonding and antibonding orbitals.
- Bond Order: 0 (indicating instability).
- Magnetic Behavior: Diamagnetic (all electrons are paired).
- (F_2^{2-}) is not a stable molecule as its bond order is zero, meaning no net bonding interaction exists between fluorine atoms.