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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

Q6(a)(ii). MO diagram of F₂; bond order and magnetism

  • 10 marks
  • Answered
Question

Draw the molecular orbital diagram for F₂ molecule, and mention its bond order and magnetic behaviour.

Answer in brief

F₂: (σ₂ₛ)²(σ^*₂ₛ)²(σ₂ₚ)²(π₂ₚ)⁴(π^*₂ₚ)⁴. Bond order =12(8-6)=1; all electrons paired, so F₂ is diamagnetic.

Key points

  • Each F atom is 2s²2p⁵, so 14 valence electrons go into the molecular orbitals made from the 2s and 2p atomic orbitals (the 1s cores do not take part).
  • Order of the levels. The 2s–2p energy gap grows across the period and is largest in fluorine (21.5 eV). The 2s and 2p sets therefore hardly mix (no s–p mixing), and σ₂ₚ lies below π₂ₚ, as in O₂. (From B₂ to N₂ mixing pushes…
  • bonding electrons N_b=2+2+4=8, antibonding Nₐ=2+4=6; σ^*₂ₚ is empty.
  • bond order=12(8-6)=1 a single F–F bond, as in the Lewis structure F–F with three lone pairs on each atom.

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. Each F atom is 2s²2p⁵, so 14 valence electrons go into the molecular orbitals made from the 2s and 2p atomic orbitals (the 1s cores do not take part).
  2. Order of the levels. The 2s–2p energy gap grows across the period and is largest in fluorine (21.5 eV). The 2s and 2p sets therefore hardly mix (no s–p mixing), and σ₂ₚ lies below π₂ₚ, as in O₂. (From B₂ to N₂ mixing pushes σ₂ₚ above π₂ₚ.)
  3. Filling. σ₂ₛ² σ^(*2)₂ₛ σ₂ₚ² π₂ₚ⁴ π^(*4)₂ₚ: bonding electrons N_b=2+2+4=8, antibonding Nₐ=2+4=6; σ^*₂ₚ is empty.
  4. bond order=12(8-6)=1 a single F–F bond, as in the Lewis structure F–F with three lone pairs on each atom.
  5. Magnetism. The two π^* orbitals hold four electrons, two in each, so there is no unpaired electron: F₂ is diamagnetic. O₂, with only two π^* electrons (one in each orbital), is paramagnetic with bond order 2.

Bond data: the F–F bond is long and weak for a Period 2 element: 141.2 pm (O₂ 120.8 pm) and about 159 kJ mol⁻¹, lower than Cl₂ (243 kJ mol⁻¹) and even Br₂ (193 kJ mol⁻¹). Four electrons in strongly antibonding π^* orbitals, the same thing as the repulsion between the lone pairs of two small atoms, weaken it. F₂ is isoelectronic with the peroxide ion O₂²⁻, also of bond order 1.

Application. the weak F–F bond and the strong bonds fluorine forms with other elements make F₂ the strongest oxidising agent among the elements. It is made only by electrolysis, of KHF₂ in anhydrous HF (Moissan, 1886) or of molten fluorides.

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Model answer to Q6(a)(ii), 2026 Paper I: solution page 36
Solution page 36 of 55