) Draw a molecular orbital energy level diagram for NF and write down the corresponding electronic configuration. In the diagram, be sure to label each molecular orbital with its name. The ionization energies of N and F are 14.5 and 17.4 eV. Note that NF is isoelectronic with O
. b) Next to the corresponding energy level, draw a picture of the LUMO. c) In the HOMO, the electron density resides more on the (circle the correct answer) N atom F atom neither, since it’s a non-bonding orbital neither, since the electron density is equally shared by N and F d) Write down the LCAO-MO wavefunction for the LUMO e) What is the bond order of NF? f) Circle the molecule(s) expected to be paramagnetic: NF NF
NF
none of them g) Circle the molecule(s) expected to have the shortest bond: NF NF
NF
h) With an arrow, indicate in the energy level diagram the lowest energy spin-allowed transition in NF. i) On the same graph, plot molecular potential energy diagrams for NF and NF
. Label the axes. Pick one of the curves and indicate R
and D
for that curve.
The Correct Answer and Explanation is:
Let’s work through the NF molecule step by step using molecular orbital (MO) theory. NF is a heteronuclear diatomic molecule with 14 electrons (same as O₂), so its MO diagram is similar to O₂, but with modifications due to differences in electronegativity and ionization energies (F has a higher electronegativity and ionization energy than N).
(a) MO Diagram and Electronic Configuration
Since NF is isoelectronic with O₂, the molecular orbitals fill as follows (energy order based on O₂-like molecules): σ1s<σ1s∗<σ2s<σ2s∗<σ2pz<π2px=π2py<π2px∗=π2py∗<σ2pz∗\sigma_{1s} < \sigma_{1s}^* < \sigma_{2s} < \sigma_{2s}^* < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*σ1s<σ1s∗<σ2s<σ2s∗<σ2pz<π2px=π2py<π2px∗=π2py∗<σ2pz∗
However, due to the heteronuclear nature and F being more electronegative, the atomic orbitals from F are lower in energy. Thus, the MOs have more F character at the lower levels and more N character at the higher levels.
Electronic configuration: (1σ)2(1σ∗)2(2σ)2(2σ∗)2(2π)4(1π∗)2(1\sigma)^2(1\sigma^*)^2(2\sigma)^2(2\sigma^*)^2(2\pi)^4(1\pi^*)^2(1σ)2(1σ∗)2(2σ)2(2σ∗)2(2π)4(1π∗)2
This totals 14 electrons.
(b) LUMO
The LUMO (Lowest Unoccupied Molecular Orbital) is the next orbital after the highest filled one. Since the last filled orbitals are the π*(2p), the LUMO is:
σ*(2p_z)
A drawing of this would show a large anti-bonding lobe along the internuclear axis with a nodal plane between the atoms.
(c) Electron Density in HOMO
The HOMO is the π*(2p) orbital.
This orbital is non-bonding, but since N is less electronegative than F, the orbital has more electron density on the N atom.
Answer: N atom
(d) LCAO-MO Wavefunction for LUMO
The LUMO is σ*(2p_z): ψσ∗=cA⋅2pz(N)−cB⋅2pz(F)\psi_{\sigma^*} = c_A \cdot 2p_z(N) – c_B \cdot 2p_z(F)ψσ∗=cA⋅2pz(N)−cB⋅2pz(F)
Where cAc_AcA and cBc_BcB are coefficients reflecting the atomic orbital contributions. The minus sign reflects antibonding character.
(e) Bond Order of NF
Bond order = (bonding electrons − antibonding electrons) ÷ 2
= (8 − 4) ÷ 2 = 2
(f) Paramagnetic Species
Look for unpaired electrons:
- NF: 14 electrons → π*(2p) filled → no unpaired electrons
- NF⁻: 15 electrons → adds to σ*(2p) → 1 unpaired electron → paramagnetic
- NF⁺: 13 electrons → 1 less in π*(2p) → 1 unpaired electron → paramagnetic
Answer: NF⁻ and NF⁺
(g) Shortest Bond
Bond order correlates with bond length.
- NF⁺ has bond order 2.5
- NF has bond order 2
- NF⁻ has bond order 1.5
Answer: NF⁺
(h) Lowest Energy Spin-Allowed Transition
From HOMO (π*(2p)) to LUMO (σ*(2p))
Draw an arrow from π*(2p) to σ*(2p) in the MO diagram.
(i) Potential Energy Curves
Sketch a graph with:
- x-axis: Internuclear distance RRR
- y-axis: Potential energy
NF⁺: Steep curve, minimum at lower R (short bond), deeper well (stronger bond)
NF⁻: Broad curve, minimum at higher R, shallower well
Label one curve with:
- ReR_eRe: equilibrium bond length (at minimum)
- DeD_eDe: bond dissociation energy (depth of well)
