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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

Q7(a)(i). Vaska's compound: structure and oxidative addition of HCl and O₂

  • 15 marks
  • Answered
Question

Write Vaska's catalyst with structure and show its oxidative addition with HCl and oxygen.

Answer in brief

Vaska's compound is trans-[IrCl(CO)(PPh₃)₂], a square-planar 16-electron Ir(I) complex. HCl adds to give octahedral Ir(III) [IrHCl₂(CO)(PPh₃)₂]; O₂ adds reversibly, side-on, to give the peroxo complex [IrCl(O₂)(CO)(PPh₃)₂].

Key points

  • Compound. trans-[IrCl(CO)(PPh₃)₂], carbonylchloridobis(triphenylphosphine)iridium(I), a yellow solid. It is made from Na₂IrCl₆ and PPh₃ in diethylene glycol under CO.
  • 8 (Ir) + 2 (Cl⁻) + 2 (CO) + 2 × 2 (PPh₃) = 16 valence electrons. The two bulky phosphines are trans; Cl and CO are trans to each other. Square planar, like other d⁸ ions of the heavy metals.
  • Oxidative addition. A molecule X–Y adds across the metal and its bond breaks: the oxidation state rises by 2 (I → III), the coordination number by 2 (4 → 6) and the electron count by 2 (16 → 18). A coordinatively unsaturated,…
  • [IrCl(CO)(PPh₃)₂] + HCl → [IrHCl₂(CO)(PPh₃)₂] Ir(III), d⁶, octahedral, 18 electrons; a hydrido–chlorido complex. H and Cl occupy the two new positions above and below the original plane (cis or trans to each other, depending…

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. Compound. trans-[IrCl(CO)(PPh₃)₂], carbonylchloridobis(triphenylphosphine)iridium(I), a yellow solid. It is made from Na₂IrCl₆ and PPh₃ in diethylene glycol under CO.
  2. Structure and electron count. Ir(I) is d⁸: 8 (Ir) + 2 (Cl⁻) + 2 (CO) + 2 × 2 (PPh₃) = 16 valence electrons. The two bulky phosphines are trans; Cl and CO are trans to each other. Square planar, like other d⁸ ions of the heavy metals.
  3. Oxidative addition. A molecule X–Y adds across the metal and its bond breaks: the oxidation state rises by 2 (I → III), the coordination number by 2 (4 → 6) and the electron count by 2 (16 → 18). A coordinatively unsaturated, electron-rich d⁸ centre favours it.
  4. With HCl: [IrCl(CO)(PPh₃)₂] + HCl → [IrHCl₂(CO)(PPh₃)₂] Ir(III), d⁶, octahedral, 18 electrons; a hydrido–chlorido complex. H and Cl occupy the two new positions above and below the original plane (cis or trans to each other, depending on the solvent and mechanism).
  5. With oxygen: [IrCl(CO)(PPh₃)₂] + O₂ ⇌ [IrCl(O₂)(CO)(PPh₃)₂] O₂ binds side-on (η²); it is reduced to a peroxo ligand O₂²-, so Ir becomes Ir(III). The O–O bond lengthens. The addition is reversible: bubbling an inert gas through the solution releases O₂ (a model for oxygen carriers).
  6. Other additions: H₂ (cis dihydride), CH₃I, Cl₂; the reverse step, reductive elimination, releases the product in catalytic cycles.

Application. oxidative addition followed by reductive elimination is the core of homogeneous catalysis: Wilkinson's catalyst [RhCl(PPh₃)₃] hydrogenates alkenes this way, and the Monsanto acetic acid process starts with oxidative addition of CH₃I to [RhI₂(CO)₂]⁻.

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Model answer to Q7(a)(i), 2017 Paper I: solution page 28
Solution page 28 of 38

All parts of Question 7