Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A4 / b / iii

Table of contents
- Solution iii

Solution

At

β_{L} = 0

, the preceding matrix separates into identical

(ϕ_{1}, ϕ_{2})

and

(ϕ_{3}, ϕ_{4})

blocks. For

β_{S} < 0

, a normalized pair at the lower energy

α + β_{S}

is

ψ_{-, 1} ψ_{-, 2} = \frac{ϕ _{1} + ϕ _{2}}{2} = \frac{p _{1} + p _{2} - p _{5} - p _{6}}{2}, = \frac{ϕ _{3} + ϕ _{4}}{2} = \frac{p _{3} + p _{4} - p _{7} - p _{8}}{2} .

A pair at the upper energy

α - β_{S}

is

ψ_{+, 1} ψ_{+, 2} = \frac{ϕ _{1} - ϕ _{2}}{2} = \frac{p _{1} - p _{2} - p _{5} + p _{6}}{2}, = \frac{ϕ _{3} - ϕ _{4}}{2} = \frac{p _{3} - p _{4} - p _{7} + p _{8}}{2} .

The four coefficient vectors are mutually orthonormal. Any orthogonal mixing within either equal-energy pair is an equally valid choice of molecular orbitals. At this limiting parameter value, additional coincidences with the one-dimensional levels occur: the full eight-orbital system is four isolated dimers, with four bonding and four antibonding levels.

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