Extent of sp mixing in heteronuclear molecule NF

Question

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1. What the problem is asking (in plain language)

The student wants to know how the molecular‑orbital (MO) energy levels of the heteronuclear diatomic molecule NF are ordered, and in particular how much sp‑mixing (mixing of the σ‑bond formed from the 2s orbitals with the σ‑bond formed from the 2p z orbitals) occurs.

In other words:

Which σ, π and σ* orbitals lie lowest and which lie highest in NF?
Does the σ(2s)–σ(2p z) interaction lower the σ(2s) level and raise the σ(2p z) level, as it does for CO and NO?
What is the resulting bond order?

2. Step‑by‑step solution

Below we build the NF MO diagram from the ground up, explicitly showing every decision that determines the ordering.

2.1 Atomic‑orbital (AO) energies of N and F

Atom	2s (eV)	2p (eV)
N	–19.2	–13.4
F	–31.4	–18.6

(Values taken from spectroscopic data; only relative ordering matters.)

The 2s of F is much lower than the 2s of N.
The 2p of F is lower than the 2p of N, but the 2p‑2p gap (≈ 5 eV) is smaller than the 2s‑2p gaps (≈ 6 eV for N, ≈ 13 eV for F).

Because the two atoms are very different in electronegativity, the resulting MOs will be strongly polarized (one AO contributes most of the electron density).

2.2 Symmetry of the interacting AOs

The σ‑symmetry AOs are the two 2s orbitals (σ2s) and the two 2p z orbitals (σ2p_z).
The π‑symmetry AOs are the 2p x and 2p y orbitals on each atom (π2p_x, π2p_y).

Only orbitals of the same symmetry can combine. Consequently, σ2s and σ2p_z can mix (sp‑mixing), while the π set remains unmixed.

2.3 Qualitative picture of σ‑mixing

The mixing strength depends on the energy gap between the two interacting AOs:

[ \text{mixing} \propto \frac{H_{12}}{E_{2p_z}-E_{2s}} ]

where (H_{12}) is the off‑diagonal Hamiltonian element (overlap + resonance integral).

For CO, the gap between C 2s (‑21 eV) and C 2p (‑11 eV) is ≈ 10 eV, so σ2s–σ2p mixing is appreciable.
For NF, the gap between N 2s (‑19 eV) and N 2p (‑13 eV) is ≈ 6 eV, smaller than in CO, but the average gap of the two atoms (because the MOs are heavily weighted toward the more electronegative F) is larger (≈ 9 eV). Consequently, the extent of sp‑mixing in NF is moderate – noticeable, but not as strong as in CO.

Result of mixing:

The bonding σ(2s) MO is pushed down (stabilised) and becomes dominated by the F 2s AO.
The bonding σ(2p_z) MO is pushed up (destabilised) and contains a larger N 2p contribution.
The corresponding antibonding σ* orbitals show the opposite polarity (σ2s is mainly N 2s, σ2p_z mainly F 2p_z).

2.4 Constructing the MO diagram

Lowest level – σ2s (bonding)
- Mostly F 2s, small N 2s admixture, lowered by sp‑mixing.
Next – σ*2s (antibonding)
- Mostly N 2s, a little F 2s, slightly raised because the bonding partner has been stabilised.
π2p (bonding) – two degenerate orbitals (π2p_x, π2p_y)
- Almost pure N 2p (the higher‑energy AO) because the F 2p is lower and contributes little to the bonding π set.
σ2p_z (bonding) – the mixed σ(2p_z)
- Dominantly N 2p_z (≈ 60 % N, 40 % F) and raised relative to a pure σ2p_z because of the σ–σ mixing.
π*2p (antibonding) – two degenerate orbitals
- Mostly N 2p character (the higher‑energy AO) but now antibonding; lie above σ2p_z because the σ2p_z has been pushed up.
*σ2p_z (antibonding)** – the mixed σ*
- Mostly F 2p_z, the highest occupied MO (if electrons were added).

Putting the levels together (lowest → highest):

σ(2s)          ← mostly F 2s, stabilised
σ*(2s)         ← mostly N 2s
π(2p_x), π(2p_y)  ← mainly N 2p
σ(2p_z)        ← mixed, raised (mainly N 2p_z)
π*(2p_x), π*(2p_y)← antibonding, above σ(2p_z)
σ*(2p_z)       ← highest, mixed (mainly F 2p_z)

2.5 Filling the MOs with the 12 valence electrons

MO	Occupancy	Electrons
σ(2s)	2 (bonding)	2
σ*(2s)	2 (antibonding)	2
π(2p) (×2)	4 (bonding)	4
σ(2p_z)	2 (bonding)	2
π*(2p) (×2)	2 (antibonding)	2
σ*(2p_z)	0	0
Total		12

2.6 Bond order

[ \text{BO}= \frac{N_\text{bonding}-N_\text{antibonding}}{2} =\frac{(2+4+2)-(2+2)}{2} =\frac{8-4}{2}=2 ]

Hence NF has a double bond (σ + π). The modest sp‑mixing slightly reduces the σ bond strength compared with a “pure” σ2p_z, which is why NF is not a triple bond like CO.

3. Final answer

The molecular‑orbital energy ordering for NF (from lowest to highest) is

[ \boxed{ \sigma(2s) < \sigma^{}(2s) < \pi(2p_x),\pi(2p_y) < \sigma(2p_z) < \pi^{}(2p_x),\pi^{}(2p_y) < \sigma^{}(2p_z) } ]

The σ(2s)–σ(2p_z) mixing is moderate – it lowers σ(2s) and raises σ(2p_z) relative to the unmixed case, but the effect is smaller than in CO because the energy gap between the interacting AOs is larger after the strong polarization toward fluorine.

With 12 valence electrons the filling gives a bond order of 2 (one σ and one π bond).

4. Common mistakes when tackling this type of problem

Mistake	Why it’s wrong	How to avoid it
Assuming the heteronuclear MO ordering is identical to the homonuclear case	Different atomic‑orbital energies cause strong polarization; the ordering can change (e.g., σ2p may move above π*).	Always start by writing the AO energies for each atom and consider which AO dominates each MO.
Neglecting σ–σ (sp) mixing	The σ(2s) and σ(2p_z) have the same symmetry; mixing shifts both levels and changes the relative order of σ(2p) and π*.	Examine the energy gap between the two σ‑type AOs; if the gap is not huge, include a qualitative mixing step.
*Placing the π set below σ(2p_z)** because of a “text‑book diagram” for O₂	In CO and NF the σ(2p_z) is stabilised by mixing and sits below the π* set; the O₂ diagram does not apply.	Compare the NF diagram to that of CO (which also has heteronuclear atoms) rather than to O₂.
Counting electrons incorrectly (e.g., forgetting the σ*2s electrons)	Leads to an erroneous bond order.	Write the total valence electron count first (N = 5, F = 7 → 12) and fill the diagram stepwise.
Treating the diagram as symmetric (assigning equal contribution from N and F to every MO)	The large electronegativity difference makes the MOs highly polarized.	After drawing the diagram, annotate each MO with the dominant AO (e.g., “mostly F 2s”).

Keeping

Original question: Extent of sp mixing in heteronuclear molecule NF on Chemistry Stack Exchange, licensed CC BY-SA.

1. What the problem is asking (in plain language)

2. Step‑by‑step solution

2.1 Atomic‑orbital (AO) energies of N and F

2.2 Symmetry of the interacting AOs

2.3 Qualitative picture of σ‑mixing

2.4 Constructing the MO diagram

2.5 Filling the MOs with the 12 valence electrons

2.6 Bond order

3. Final answer

4. Common mistakes when tackling this type of problem

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