The successive ionization enthalpies (in kJ mol^ -1 ) of an element...

The successive ionization enthalpies of element M are given as:
$IE_1 = 577 \text{ kJ mol}^{-1}$
$IE_2 = 1816 \text{ kJ mol}^{-1}$
$IE_3 = 2744 \text{ kJ mol}^{-1}$
$IE_4 = 11577 \text{ kJ mol}^{-1}$
$IE_5 = 14842 \text{ kJ mol}^{-1}$

There is a sudden large jump in the ionization enthalpy between $IE_3$ and $IE_4$. This indicates that the removal of the fourth electron requires a very high amount of energy, which means the fourth electron is being removed from a stable noble gas core.

Thus, element M has 3 valence electrons and its most stable oxidation state is $+3$.

The formula of the oxide formed by an element in the $+3$ oxidation state is $M_2O_3$.

Answer: $M_{2}O_{3}$