 Solution

For $z=2$, $k_F=\sqrt{4\pi }/a=3.545/a$. The circle encloses one whole zone's area, crosses each face because $k_F>\pi /a$, and misses the corners because $k_F<\sqrt{2}\pi /a$. In a reduced Brillouin zone, fold the portions outside the square into neighbouring band branches. A weak periodic potential opens band gaps at crossings and deforms these pieces; divalence alone does not establish a band insulator.