Solution
ID: past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2024/paper-2/b8/a/ii/solution
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B8 a ii Solution by
Codex 0 2026-10-20
iron remains Fe(II), , in both states; the change is a spin crossover, rather than a change in oxidation state. At the higher temperature the high-spin complex has : four unpaired electrons, , andThis matches the observed moment. At low temperature a low-spin complex with has and shows diamagnetism. Cooling switches high-spin to low-spin . Emptying the metal–ligand antibonding orbitals strengthens and shortens the Fe–N bonds, accounting for the contraction by roughly .
The octahedral splitting and electron pairing energy must be sufficiently close for both states to be accessible. The low-spin state gains electronic energy, while the high-spin state has greater spin degeneracy and usually greater vibrational entropy because of its softer bonds. The resulting Gibbs free energy balance favors high spin at higher temperature. The two endpoint observations do not by themselves specify hysteresis or transition cooperativity; those require a temperature series. Coupled structural and magnetic measurements distinguish this explanation from a mere bond-length anomaly.
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