Fill four spatial molecular orbitals with the eight

π

electrons. For negative

β_{S}, β_{L}

and

∣ β_{S} ∣ > ∣ β_{L} ∣

, the occupied energies are

α + β_{S} + β_{L}

α + β_{S} - β_{L}

, and the pair at

α - β_{S}^{2} + β_{L}^{2}

. Therefore

E_{π} = 8 α + 4 β_{S} - 4 β_{S}^{2} + β_{L}^{2} .

For equal integrals,

β_{S} = β_{L} = β < 0

, this becomes

E_{π}^{uniform} = 8 α + 4 (1 + 2) β .

At equal bond lengths, the last two electrons occupy the frontier degeneracy at

E = α

, so their distribution within that degenerate space does not change the bare Hückel method energy.

2

Solution

For

β_{L} = 0

, four isolated short-bond dimers each contribute

2 (α + β_{S})

E_{π}^{dimers} = 8 α + 8 β_{S} .

These are different coupling limits. Keeping

β_{S} = β

while simply removing long-bond coupling does not represent a small distortion at fixed mean bond strength.

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