= Solution
For divalent atoms, $k_Fa=\sqrt{4\pi}$. The circle has the same area as the <first Brillouin zone> but crosses its faces and leaves its corners unfilled. Thus \b[the free-electron case remains conducting]: filling capacity alone does not remove band overlap. A nearly free-electron conductor has reconstructed <electron> and <hole pockets>; a divalent <band insulator> instead has one completely filled band separated from all higher bands, with no <Fermi surface> in the gap.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2014/paper-2-square-fermi.png]
{title=Original square-zone state counting, monovalent and divalent free contours, and insulating band filling}
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