 Solution

ID: 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/a/solution

For positive

Δ_{o} ≫ B

, expand the common radical as

Δ_{o}^{2} + 18 B Δ_{o} + 225 B^{2} = Δ_{o} + 9 B + \frac{72 B ^{2}}{Δ _{o}} + O (\frac{B ^{3}}{Δ _{o}^{2}}) .

Substitution into the stated crystal field theory energies gives

E_{0} E_{1} E_{2} E_{3} = - \frac{4}{5} Δ_{o} + 3 B - \frac{36 B ^{2}}{Δ _{o}} + \dots, = \frac{1}{5} Δ_{o}, = \frac{1}{5} Δ_{o} + 12 B + \frac{36 B ^{2}}{Δ _{o}} + \dots, = \frac{6}{5} Δ_{o} .

The leading straight-line asymptotes retain the terms linear in

Δ_{o}

and constant in

B

. The

B^{2} / Δ_{o}

terms show from which side the interacting

T_{1 g}

levels approach them.

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