Use molecular orbital theory to complete the ground state electron configuration for each of the molecules. Molecule NF NF+ NF NF bond order: Ground state electron configuration NF+ bond order: Determine the bond order of each of the molecules. Answer Bank 3 Question source: McQuarrie, Rock, And Gallogly 4e – General Chemistry
The Correct Answer and Explanation is:
To solve this problem using molecular orbital (MO) theory, we will examine the molecular orbital configurations of NF and NF⁺ and determine their bond orders.
Step 1: Determine the total number of electrons
- N (Nitrogen) has 7 electrons
- F (Fluorine) has 9 electrons
- So, NF has 7 + 9 = 16 electrons
- NF⁺ loses one electron → 15 electrons
Step 2: Use the MO energy diagram for heteronuclear diatomic molecules
NF is a heteronuclear diatomic molecule. The molecular orbital ordering resembles that of oxygen and fluorine molecules (because F dominates in electronegativity). The orbital energy order is:
scssCopyEditσ(1s), σ*(1s), σ(2s), σ*(2s), σ(2p), π(2p), π*(2p), σ*(2p)
Step 3: Fill the molecular orbitals
For NF (16 electrons):
Fill the orbitals in order:
- σ(1s)²
- σ*(1s)²
- σ(2s)²
- σ*(2s)²
- σ(2p)²
- π(2p)⁴
- π*(2p)²
Bond Order = ½ × [(# of bonding electrons) − (# of antibonding electrons)]
= ½ × [(2 + 2 + 2 + 4) − (2 + 2 + 2)]
= ½ × [10 − 6]
= 2
For NF⁺ (15 electrons):
Remove one electron from the highest energy orbital (π*(2p)):
- σ(1s)²
- σ*(1s)²
- σ(2s)²
- σ*(2s)²
- σ(2p)²
- π(2p)⁴
- π*(2p)¹
Bond Order = ½ × [(10) − (5)] = 2.5
Final Answers:
- NF bond order: 2
- NF⁺ bond order: 2.5
Ground State Electron Configurations:
NF (16e⁻):
σ(1s)² σ*(1s)² σ(2s)² σ*(2s)² σ(2p)² π(2p)⁴ π*(2p)²
NF⁺ (15e⁻):
σ(1s)² σ*(1s)² σ(2s)² σ*(2s)² σ(2p)² π(2p)⁴ π*(2p)¹
Explanation:
Molecular orbital theory combines atomic orbitals to form molecular orbitals that are spread over the entire molecule. Electrons fill lower-energy bonding orbitals before entering higher-energy antibonding orbitals. A bond order reflects the net bonding strength between atoms: the higher the bond order, the stronger the bond. In this case, NF⁺ has a higher bond order than NF, which means removing an electron (from an antibonding orbital) actually strengthens the bond. This is a good illustration of how molecular orbital theory can predict trends in bond strength based on electron configuration.
