Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 2A / Section A / 16

For positive ${\Delta }_o\gg B$, expand the common radical asSubstitution into the stated crystal field theory energies givesThe leading straight-line asymptotes retain the terms linear in ${\Delta }_o$ and constant in $B$. The $B^2/{\Delta }_o$ terms show from which side the interacting $T_{1g}$ levels approach them.

For $B\gg {\Delta }_o$, the corresponding asymptotic expansion isHenceAt zero field, $E_0,E_1,E_3$ converge to the ${}^{3}F$ atomic term and $E_2$ to ${}^{3}P$ at $15B$. The ${}^3{T}_{1g}(P)$ branch therefore has zero initial slope; its high-field slope is not zero.

The descending line is ${}^3{T}_{1g}(F)$; the low rising line is ${}^3{T}_{2g}$; the steep line is ${}^3{A}_{2g}$; and the branch starting at $15B$ is ${}^3{T}_{1g}(P)$. Their high-field slopes, in the same order, areThese are slopes of the supplied absolute term energies relative to their chosen reference, not slopes of transition energies measured upward from the ground state. The two $T_{1g}$ states share a symmetry and mix, whereas the crossing of $A_{2g}$ and the upper $T_{1g}$ branch is allowed because they have different irreducible representations.