 Solution

The extra electron occupies the lowest unoccupied molecular orbital of the neutral molecule, the $B_{3g}$ orbital at the middle root ${\lambda }_L=0.25410169\dots$ of ${\lambda }^3-4\lambda +1=0$. WriteThe first and third rows of the $B_{3g}$ block giveChoosing 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,The squared coefficients sum to one. Within this fixed-orbital Hückel orbital charge density approximation, the added electron contributes site populationsThus 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.