The successive 5 ionisation energies of an element are 800 , 2427 , 3658 , 25024 and 32824 kJ / mol , respectively. By using the above values predict the group in which the above element is present :

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The successive 5 ionisation energies of an element are $800, 2427, 3658, 25024$ and $32824 kJ / mol$ , respectively.
$\frac{IE _{2}}{IE _{1}} = \frac{2427}{800} = 3.03 \frac{IE}{IE _{2}} = \frac{3658}{2427} = 1.51 \frac{IE}{IE _{3}} = \frac{25024}{3658} = 6.84$
$\frac{IE _{5}}{IE _{4}} = \frac{32824}{25024} = 1.31$
Since ( $IE_{4} / IE_{3}$ ) value is maximum, the element belongs to group 13.

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