Q2(a)(i). MO diagram of NO; NO⁺, CN⁻ and CO; bond order of NO and NO⁺
Draw the molecular orbital diagram of NO molecule. Show that NO⁺, CN⁻ and CO are isoelectronic, and the bond order of NO is lower than that of NO⁺.
Answer in brief
NO (11 valence electrons): (σ₂ₛ)²(σ*₂ₛ)²(σ₂ₚ)²(π₂ₚ)⁴(π*₂ₚ)¹, bond order 2.5. NO⁺, CN⁻ and CO each have 14 electrons (10 valence), bond order 3. Ionisation of NO removes the antibonding π* electron, so the bond order rises from 2.5 to 3.
Key points
- The 2s and 2p atomic orbitals of N and O combine. O is more electronegative, so its atomic orbitals lie lower; the bonding MOs are closer to O in energy and character, the antibonding MOs closer to N.
- NO has 5+6=11 valence electrons (15 in all). Filling in order of energy: 3pt] (σ₂ₛ)²(σ*₂ₛ)²(σ₂ₚ)²(π₂ₚ)⁴(π*₂ₚ)¹ 3pt] Some texts put σ₂ₚ just above π₂ₚ (as in N₂); the odd electron is in π*₂ₚ either way.
- b.o.(NO)=12(8-3)=2.5
- NO is paramagnetic, an odd-electron molecule (a stable radical).
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
The diagram
- The 2s and 2p atomic orbitals of N and O combine. O is more electronegative, so its atomic orbitals lie lower; the bonding MOs are closer to O in energy and character, the antibonding MOs closer to N.
- NO has 5+6=11 valence electrons (15 in all). Filling in order of energy: 3pt] (σ₂ₛ)²(σ*₂ₛ)²(σ₂ₚ)²(π₂ₚ)⁴(π*₂ₚ)¹ 3pt] Some texts put σ₂ₚ just above π₂ₚ (as in N₂); the odd electron is in π*₂ₚ either way.
- Bonding electrons N_b=8, antibonding Nₐ=3: b.o.(NO)=12(8-3)=2.5
- One unpaired electron: NO is paramagnetic, an odd-electron molecule (a stable radical).
labblue NO⁺, CN⁻ and CO are isoelectronic
| Species | Electron count | Total e | Valence e | Bond order |
| NO⁺ | 7+8-1 | 14 | 10 | 3 |
| CN⁻ | 6+7+1 | 14 | 10 | 3 |
| CO | 6+8 | 14 | 10 | 3 |
| (N₂, for comparison) | 7+7 | 14 | 10 | 3 |
Same number of electrons in the same set of MOs: all three have (σ₂ₛ)²(σ*₂ₛ)²(π₂ₚ)⁴(σ₂ₚ)², a triple bond (σ + 2π), no unpaired electron, diamagnetic. They are isoelectronic with N₂.
Why the bond order of NO is lower than that of NO⁺
- Ionisation removes the electron of highest energy, the one in the antibonding π*₂ₚ orbital: NO → NO⁺ + e⁻
- Losing an antibonding electron strengthens the bond: b.o.(NO⁺)=12(8-2)=3 > b.o.(NO)=2.5
- The measured properties agree: bond length 115 pm (NO) → 106 pm (NO⁺); bond dissociation energy 626 → 1047 kJ mol⁻¹; N–O stretching wavenumber 1876 → about 2300 cm⁻¹.
Application. NO⁺ (the nitrosonium ion) is a stable cation, as in NO+BF₄⁻ and NO+HSO₄⁻. In metal nitrosyls a linear M–N–O unit is counted as NO⁺, isoelectronic with CO, as in nitroprusside [Fe(CN)₅NO]²⁻; CO and CN⁻, its isoelectronic partners, are the classic π-acceptor ligands.
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All parts of Question 2
- 2(a)(i) MO diagram of NO; NO⁺, CN⁻ and CO; bond order of NO and NO⁺15 marks
- 2(a)(ii) Theory of absolute reaction rates; how a catalyst works10 marks
- 2(b) Integrated Clausius–Clapeyron equation from the Clapeyron equation15 marks
- 2(c) Heat of reaction for N₂ + 3H₂ → 2NH₃ at 50°C: Kirchhoff's law10 marks