Q3(b). cis–trans isomerisation of PtCl₂(PEt₃)₂
cis-PtCl₂(PEt₃)₂ is stable in benzene solution. However, a small amount of free triethylphosphine catalyzes the establishment of an equilibrium with the trans-isomer: cis-PtCl₂(PEt₃)₂ ⇌ trans-PtCl₂(PEt₃)₂. For the conversion of cis to trans in benzene at 25°C, Δ H^°=10.3 kJ mol⁻¹ and Δ S^°=55.6 J mol⁻¹ K⁻¹. (i) Calculate the equilibrium constant. (ii) Which isomer has higher bond energy? Explain briefly. (iii) Why is free triethylphosphine necessary to catalyze the isomerization?
Answer in brief
(i) K=12.6 (Δ G^°=-6.28 kJ mol⁻¹). (ii) The cis isomer (larger positive Δ H to reach trans). (iii) Free PEt₃ gives a five-coordinate intermediate, an associative path that allows the ligands to rearrange.
Key points
- Square-planar d⁸ Pt(II) substitutes by the associative path (16e→18e); simple intramolecular twisting of the planar complex has a very high barrier.
- PEt₃ attacks to give five-coordinate [PtCl₂(PEt₃)₃] (trigonal bipyramid, 18e). The ligands exchange positions by pseudorotation, and loss of PEt₃ gives either isomer.
- With no free phosphine, nothing starts this path and the cis form is kinetically inert; the equilibrium is reached only when a little PEt₃ is added: it is a catalyst because it is released again.
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
(i) Equilibrium constant
Δ G^°=Δ H^°-TΔ S^°=10300-(298.15)(55.6)=10300-16577=-6277 J mol⁻¹
ln K=-(Δ G^°)/(RT)=(6277)/((8.314)(298.15))=2.532 K=12.6
Trans is favoured by entropy: at 298 K, TΔ S^°=16.6 kJ exceeds Δ H^°=10.3 kJ; the two are equal at T=Δ H^°/Δ S^°=185 K. At 25°C about 93% is trans.
(ii) Bond energy
Δ H^°>0 for cis → trans, so bonds are stronger in the cis isomer. In cis, each PEt₃ is trans to Cl⁻; in trans, the two strongly σ-donating (high trans-influence) PEt₃ compete for the same Pt d orbital (pₓ/d_(x²-y²)) and weaken each other's Pt–P bond. Cis also benefits from π back-donation to P from two separate d orbitals. Trans is favoured by entropy and steric relief between the bulky PEt₃ groups (cone angle 132°).
(iii) Role of free phosphine
- Square-planar d⁸ Pt(II) substitutes by the associative path (16e→18e); simple intramolecular twisting of the planar complex has a very high barrier.
- PEt₃ attacks to give five-coordinate [PtCl₂(PEt₃)₃] (trigonal bipyramid, 18e). The ligands exchange positions by pseudorotation, and loss of PEt₃ gives either isomer.
- With no free phosphine, nothing starts this path and the cis form is kinetically inert; the equilibrium is reached only when a little PEt₃ is added: it is a catalyst because it is released again.
the same associative route, catalysed by traces of a nucleophile, explains cis–trans isomerisation of many square-planar Pt(II) complexes (Basolo and Pearson).
Coversthermodynamics and mechanismcis–trans isomerisation of PtCl₂(PEt₃)₂
All parts of Question 3
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