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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 · 2026 · Paper I

Q5(i). Borazine, the “inorganic benzene”

  • 5 marks
  • Answered
Question

What is inorganic benzene? Explain it with structure.

Answer in brief

Borazine, B₃N₃H₆: a planar six-membered ring of alternating B and N atoms, isoelectronic and isostructural with benzene.

Key points

  • heating diborane with ammonia, 3B₂H₆ + 6NH₃ → 2B₃N₃H₆ + 12H₂ (Stock, 1926); or 3NH₄Cl + 3BCl₃ → B₃N₃H₃Cl₃ + 9HCl, followed by LiBH₄.
  • a B–N pair has the same number of electrons as a C–C pair, so B₃N₃H₆ is isoelectronic with C₆H₆. Both are planar hexagons with equal ring bonds and 120° angles; densities 0.81 (borazine) and 0.88 g cm⁻³ (benzene); μ=0 for both.
  • each B and N is sp². Each N gives its lone pair to the π system and each B has an empty 2p orbital, so six π electrons spread round the ring; all B–N 144 pm.
  • N is much more electronegative than B, so the π density stays mainly on N (N δ-, B δ+). The delocalisation is only partial and the ring bonds are polar.

The full model answer

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

  1. Preparation: heating diborane with ammonia, 3B₂H₆ + 6NH₃ → 2B₃N₃H₆ + 12H₂ (Stock, 1926); or 3NH₄Cl + 3BCl₃ → B₃N₃H₃Cl₃ + 9HCl, followed by LiBH₄.
  2. Why “inorganic benzene”: a B–N pair has the same number of electrons as a C–C pair, so B₃N₃H₆ is isoelectronic with C₆H₆. Both are planar hexagons with equal ring bonds and 120° angles; densities 0.81 (borazine) and 0.88 g cm⁻³ (benzene); μ=0 for both.
  3. Bonding: each B and N is sp². Each N gives its lone pair to the π system and each B has an empty 2p orbital, so six π electrons spread round the ring; all B–N 144 pm.
  4. Difference from benzene: N is much more electronegative than B, so the π density stays mainly on N (N δ-, B δ+). The delocalisation is only partial and the ring bonds are polar.
  5. Hence borazine adds 3HCl across the B–N bonds (H to N, Cl to B) to give B₃N₃H₉Cl₃; benzene does not.

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Model answer to Q5(i), 2026 Paper I: solution page 33
Solution page 33 of 55
Read the whole 2026 solution2026 question paper

All parts of Question 5