\b[The <ground state> changes from high-spin $t_{2g}^3e_g^2$, ${}^6A_{1g}$, to low-spin $t_{2g}^5$, ${}^2T_{2g}$.] The larger splitting eventually compensates the additional pairing cost. Because all plotted excitation energies are re-referenced to the new <ground state>, the branches change slope at the crossover. The break is not an infinite physical excitation energy or a discontinuity of the underlying <Hamiltonian>.

Solution

Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section A / 15 / b / Solution

 Solution (source code)