Chemistry - PERIODIC TABLE Question with Solution | TestHub

The successive 5 ionisation energies of an element are $800,2427,3658,25024$ and $32824\mathrm{kJ}/\mathrm{mol}$, respectively. $\begin{array}{rl}& \frac{{\mathrm{IE}}_2}{{\mathrm{IE}}_1}=\frac{2427}{800}=3.03\\ & \frac{\mathrm{IE}}{{\mathrm{IE}}_2}=\frac{3658}{2427}=1.51\\ & \frac{\mathrm{IE}}{{\mathrm{IE}}_3}=\frac{25024}{3658}=6.84\end{array}$ $\frac{{\mathrm{IE}}_5}{{\mathrm{IE}}_4}=\frac{32824}{25024}=1.31$ Since ( ${\mathrm{IE}}_4/{\mathrm{IE}}_3$ ) value is maximum, the element belongs to group 13.