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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 · 2018 · Paper I

Q7(d). Two isomers of [PtBrCl(NH₃)(py)] from [PtCl₄]²⁻ by the trans effect

  • 10 marks
  • Answered
Question

How do you prepare the above isomers starting from [PtCl₄]²⁻ using the concept of trans effect?
(Order of trans directing ability: Br⁻ > Cl⁻ > py > NH₃)
q7d_q

Answer in brief

Isomer I (Br trans to NH₃, Cl trans to py): add Br⁻, then NH₃, then py. Isomer II (Br trans to py, Cl trans to NH₃): add NH₃, then Br⁻, then py.

Key points

  • in substitution of square-planar Pt(II), the incoming ligand replaces the group trans to the ligand of highest trans effect. The order of entry is therefore chosen so that each new ligand lands in the right position.
  • [PtCl₄]²⁻ + Br⁻ → [PtBrCl₃]²⁻ + Cl⁻ (all four Cl equivalent).
  • [PtBrCl₃]²⁻ + NH₃ → [PtBrCl₂(NH₃)]⁻ + Cl⁻. Br⁻ is the strongest trans director, so the Cl trans to Br is replaced: NH₃ enters trans to Br.
  • [PtBrCl₂(NH₃)]⁻ + py → [PtBrCl(NH₃)(py)] + Cl⁻. The two remaining Cl are trans to each other (Cl trans director); one is replaced by py, which ends up trans to Cl. Product: Br trans NH₃, Cl trans py = isomer I.

The full model answer

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

Reading the figure: in each drawing the two diagonals through Pt are the trans pairs. Isomer I: Br trans to NH₃, Cl trans to py. Isomer II: Br trans to py, Cl trans to NH₃.

  1. Trans effect: in substitution of square-planar Pt(II), the incoming ligand replaces the group trans to the ligand of highest trans effect. The order of entry is therefore chosen so that each new ligand lands in the right position.

Isomer I: Br trans to NH₃, Cl trans to py

  1. [PtCl₄]²⁻ + Br⁻ → [PtBrCl₃]²⁻ + Cl⁻ (all four Cl equivalent).
  2. [PtBrCl₃]²⁻ + NH₃ → [PtBrCl₂(NH₃)]⁻ + Cl⁻. Br⁻ is the strongest trans director, so the Cl trans to Br is replaced: NH₃ enters trans to Br.
  3. [PtBrCl₂(NH₃)]⁻ + py → [PtBrCl(NH₃)(py)] + Cl⁻. The two remaining Cl are trans to each other (Cl trans director); one is replaced by py, which ends up trans to Cl. Product: Br trans NH₃, Cl trans py = isomer I.

Isomer II: Br trans to py, Cl trans to NH₃

  1. [PtCl₄]²⁻ + NH₃ → [PtCl₃(NH₃)]⁻ + Cl⁻.
  2. [PtCl₃(NH₃)]⁻ + Br⁻ → [PtBrCl₂(NH₃)]⁻ + Cl⁻. The two Cl trans to each other are labilised by Cl; the Cl trans to the weak director NH₃ is not. Br enters cis to NH₃, trans to Cl.
  3. [PtBrCl₂(NH₃)]⁻ + py → [PtBrCl(NH₃)(py)] + Cl⁻. Now the Cl trans to Br is the most labile; py replaces it and lands trans to Br. The Cl left is trans to NH₃. Product = isomer II.
  1. Isomer II can also be made by Br⁻, then py (enters trans to Br), then NH₃.
  2. The trans effect is kinetic: it arises from σ-donation weakening the trans bond (ground state) and from π-acceptance stabilising the five-coordinate transition state.

Application. the same principle is used to make the anticancer drug cisplatin, cis-[PtCl₂(NH₃)₂]: from [PtCl₄]²⁻ and NH₃ (the second NH₃ enters trans to Cl), while [Pt(NH₃)₄]²⁺ and Cl⁻ give the inactive trans isomer.

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Model answer to Q7(d), 2018 Paper I: solution page 46
Solution page 46 of 55
Read the whole 2018 solution2018 question paper

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