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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2017 · Paper I

Q4(b). Paramagnetism of NO from its MO diagram

  • 10 marks
  • Answered
Question

Explain the paramagnetism of NO with its molecular orbital diagram.

Answer in brief

NO has 11 valence electrons; the last one occupies a π*2p orbital alone. One unpaired electron makes NO paramagnetic (μₛ=1.73 μ_B); bond order 2.5.

Key points

  • NO is a heteronuclear diatomic. O is more electronegative, so its atomic orbitals lie lower; the bonding MOs have more O character, the antibonding MOs more N character.
  • Configuration (11 valence electrons, 15 in all): (σ2s)²(σ*2s)²(π2p)⁴(σ2p)²(π*2p)¹. N and O are close in the period, so the π2p level lies near (here below) σ2p; either order gives the same result.
  • the eleventh electron enters one of the two degenerate π*2p orbitals and has no partner. NO is an odd-electron molecule, so it is paramagnetic: μₛ=√(n(n+2))=√3= 1.73 μ_B
  • Bond order =12(8-3)= 2.5. Bond length 115 pm.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

VALENCE e5 (N) + 6 (O) = 11UNPAIRED1 in π*BOND ORDER2.5

  1. NO is a heteronuclear diatomic. O is more electronegative, so its atomic orbitals lie lower; the bonding MOs have more O character, the antibonding MOs more N character.
  2. Configuration (11 valence electrons, 15 in all): (σ2s)²(σ*2s)²(π2p)⁴(σ2p)²(π*2p)¹. N and O are close in the period, so the π2p level lies near (here below) σ2p; either order gives the same result.
  3. Unpaired electron: the eleventh electron enters one of the two degenerate π*2p orbitals and has no partner. NO is an odd-electron molecule, so it is paramagnetic: μₛ=√(n(n+2))=√3= 1.73 μ_B
  4. Bond order =12(8-3)= 2.5. Bond length 115 pm.
  5. Check by ionisation. Removing the π* electron gives NO⁺ (10 valence electrons, isoelectronic with N₂ and CO): bond order 3, diamagnetic, shorter bond (106 pm). NO⁻ would have 12 electrons, bond order 2, two unpaired electrons (like O₂).
  6. Because the odd electron lies in an antibonding orbital that is mainly on N, NO loses it easily (low ionisation energy) and bonds to metals through N as a ligand.

Application. NO is a biological messenger (vasodilator); its odd electron lets it bind to the iron of guanylate cyclase and to haem, and the sodium nitroprusside used in hypertension releases NO.

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Model answer to Q4(b), 2017 Paper I: solution page 16
Solution page 16 of 38
Read the whole 2017 solution2017 question paper

All parts of Question 4