The successive ionization energies of an element $\mathrm{A}$ are ${\mathrm{IE}}_1=6.0 \mathrm{eV}, {\mathrm{IE}}_2=18.8 \mathrm{eV}, {\mathrm{IE}}_3=30.2 \mathrm{eV}$ and ${\mathrm{IE}}_4=120 \mathrm{eV}$ formula of oxide of element $\mathrm{A}$ is

The formation of the oxide ion ${\mathrm{O}}^{2-} \left(\mathrm{g}\right)$ requires first, an exothermic and then an endothermic step as shown below:

$\mathrm{O} \left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{-} \left(\mathrm{g}\right); \mathrm{\Delta H}=-142 \frac{\mathrm{kJ}}{\mathrm{mol}}$,

${\mathrm{O}}^{-} \left(\mathrm{g}\right)+\mathrm{e}\to {\mathrm{O}}^{2-} \left(\mathrm{g}\right); \mathrm{\Delta H}=844 \frac{\mathrm{kJ}}{\mathrm{mol}}$.

This is because:

The electron gain enthalpies $\left(\mathrm{in} \frac{\mathrm{kJ}}{\mathrm{mol}}\right)$ of halogens are as given below:

$\mathrm{F}=-332, \mathrm{Cl}=-349, \mathrm{Br}=-324, \mathrm{I}=-295.$

The less negative value for $\mathrm{F}$ as compared to that of $\mathrm{Cl}$ is due to