Q6(c)(i). Lanthanoid–EDTA formation enthalpy
Discuss the following observation: (i) ΔH° for the formation of [Ln(EDTA)(OH₂)ₓ]⁻ (x = 2 or 3) in aqueous solution is nearly constant for all Ln and is almost zero.
Answer in brief
Largely electrostatic Ln–donor bonding and dehydration contributions approximately compensate; favourable entropy can dominate complex formation.
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Coverslanthanide EDTA complexesenthalpy of complexationhydrationentropy-driven chelationlanthanoid contraction
All parts of Question 6
- 6(a) Thermal decomposition of calcium oxalate15 marks
- 6(b) Organic precipitating reagents5 marks
- 6(c)(i) Lanthanoid–EDTA formation enthalpy5 marks
- 6(c)(ii) Non-stoichiometry of f-block oxides5 marks