= Solution
The extra electron occupies the <lowest unoccupied molecular orbital> of the neutral molecule, the $B_{3g}$ orbital at the middle root $\lambda_L=0.25410169\ldots$ of $\lambda^3-4\lambda+1=0$. Write
$$
\psi_L=aO_-+bC_-+cR_-.
$$
The first and third rows of the $B_{3g}$ block give
$$
b=(\lambda_L-1)a,\qquad c=\frac{\sqrt2(\lambda_L-1)}{1+\lambda_L}a,\qquad
a=\left[1+(\lambda_L-1)^2+\frac{2(\lambda_L-1)^2}{(1+\lambda_L)^2}\right]^{-1/2}.
$$
Choosing the overall phase so that site 1 is positive gives $(a,b,c)\simeq(0.664623,-0.495741,-0.559033)$. In the original site order,
$$
\boxed{\psi_L\simeq0.469959p_1-0.350542p_2-0.279516p_3-0.279516p_4
+0.279516p_5+0.279516p_6+0.350542p_7-0.469959p_8.}
$$
The squared coefficients sum to one. Within this fixed-orbital <Hückel orbital charge density> approximation, the added electron contributes site populations
$$
\Delta n_{1,8}\simeq0.220862\ \text{each},\qquad
\Delta n_{2,7}\simeq0.122880\ \text{each},\qquad
\Delta n_{3,4,5,6}\simeq0.078129\ \text{each}.
$$
Thus the oxygen sites $\boxed{1\text{ and }8}$ receive the largest added negative charge, each receiving an increment of approximately $-0.221e$. Including the four doubly occupied neutral orbitals gives an anion $\pi$ population of about $1.621$ on each oxygen, $0.870$ on sites 2 and 7, and $1.005$ on each remaining carbon. Thus oxygen also has the largest total $\pi$ population in this model. A coefficient's sign is an orbital phase, not the sign of its electric charge. Predicting an actual charge density would additionally require orbital relaxation and a population-analysis convention; the stated result is the requested orbital model.
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