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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 · 2019 · Paper I

Q7(c). Why N₂⁺ is weaker but O₂⁺ stronger than the neutral molecule

  • 10 marks
  • Answered
Question

When an electron is removed from an N₂ molecule, forming N₂⁺ ion, the bond between the N atoms is weakened. When an O₂ molecule is similarly ionized to O₂⁺, the bond between O atoms is strengthened. Explain this difference.

Answer in brief

The electron leaves the HOMO. In N₂ the HOMO (σ₂ₚ) is bonding, so the bond order falls from 3 to 2.5; in O₂ the HOMO (π*₂ₚ) is antibonding, so the bond order rises from 2 to 2.5.

Key points

  • σ₂ₛ²σ₂ₛ*²π₂ₚ⁴σ₂ₚ². s–p mixing pushes σ₂ₚ above π₂ₚ, so the HOMO is the bonding σ₂ₚ. Bond order 12(8-2)=3.
  • one bonding electron lost, σ₂ₚ¹; bond order 12(7-2)=2.5. The bond is longer and weaker: 109.8 pm → 111.6 pm.
  • σ₂ₛ²σ₂ₛ*²σ₂ₚ²π₂ₚ⁴π₂ₚ*². The HOMO is the antibonding π*₂ₚ (two unpaired electrons). Bond order 12(8-4)=2.
  • one antibonding electron lost, π₂ₚ*¹; bond order 12(8-3)=2.5. The bond is shorter and stronger: 120.8 pm → 111.6 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.

  1. N₂ (10 valence electrons): σ₂ₛ²σ₂ₛ*²π₂ₚ⁴σ₂ₚ². s–p mixing pushes σ₂ₚ above π₂ₚ, so the HOMO is the bonding σ₂ₚ. Bond order 12(8-2)=3.
  2. N₂⁺: one bonding electron lost, σ₂ₚ¹; bond order 12(7-2)=2.5. The bond is longer and weaker: 109.8 pm → 111.6 pm.
  3. O₂ (12 valence electrons): σ₂ₛ²σ₂ₛ*²σ₂ₚ²π₂ₚ⁴π₂ₚ*². The HOMO is the antibonding π*₂ₚ (two unpaired electrons). Bond order 12(8-4)=2.
  4. O₂⁺: one antibonding electron lost, π₂ₚ*¹; bond order 12(8-3)=2.5. The bond is shorter and stronger: 120.8 pm → 111.6 pm.
  5. Magnetism: N₂ is diamagnetic, N₂⁺ has one unpaired electron; O₂ has two, O₂⁺ one. Ionisation thus changes bond strength in opposite directions because it empties orbitals of opposite character.

Application. the strong bond of the dioxygenyl cation allowed Bartlett to make O₂+[PtF₆]⁻ from O₂ and PtF₆; since the first ionisation energies of O₂ and Xe are almost equal, he then made the first xenon compound.

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Model answer to Q7(c), 2019 Paper I: solution page 33
Solution page 33 of 40
Read the whole 2019 solution2019 question paper

All parts of Question 7