Find the formula of halide of a metal whose successive ionization energies are 10, 16, 25, 400, 500 eV mol$^{-1}$ respectively:

The large jump in ionization energy occurs between the 3rd and 4th ionizations: $25 \to 400$. This means the atom loses three valence electrons relatively easily, but removing the fourth would disturb a stable inner shell. Hence the metal commonly forms $M^{3+}$, so its halide is $MX_3$.