For an element (x) the successive ionization energies ${\mathrm{IE}}_1,{\mathrm{IE}}_2,{\mathrm{IE}}_3,{\mathrm{IE}}_4$ and ${\mathrm{IE}}_5$  are $800, 2427, 3658$ , $25024$  and $32824 \mathrm{kJ} {\mathrm{mol}}^{-1}$ respectively, then what is  the number of valence  electrons present  in the  element ?

III A gp elements : ${\mathrm{IE}}_1<{\mathrm{IE}}_2<{\mathrm{IE}}_3<<<{\mathrm{IE}}_4<{\mathrm{IE}}_5$ ${\mathrm{ns}}^2{\mathrm{np}}^1$$\therefore$ no of valence $\overline{\mathrm{e}}\mathrm{s}=3$.

We can that the fourth value of IE is too high, it means that removal of this electron needs lots of energy. Hence we can say removal of three electrons is easy.