= Solution
The <first Brillouin zone> is $|k_x|,|k_y|\le\pi/a$. For one atom per primitive cell and two spin states, a monovalent free-electron gas has
$$
n_{2D}=\frac1{a^2}=\frac{k_F^2}{2\pi},\qquad
\boxed{k_Fa=\sqrt{2\pi}}.
$$
Its circular <Fermi surface> lies inside the zone and encloses half its area: \b[a partially filled band is conducting]. Here <odd filling obstructs a spin-degenerate band insulator>: a monovalent <band insulator> is impossible in this spin-degenerate one-atom independent-electron model. It would require changed premises, such as unit-cell enlargement or physics beyond the free-electron picture; no insulating free-electron contour should be invented.
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