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:

We are given five successive ionization energies of an element: 800, 2427, 3658, 25024, and 32824 kJ/mol. We need to predict the group of this element.

The key to determining the group is finding where a dramatic increase occurs between consecutive ionization energies:

- IE$$_1$$ = 800 kJ/mol

- IE$$_2$$ = 2427 kJ/mol (ratio to IE$$_1$$ = 3.0)

- IE$$_3$$ = 3658 kJ/mol (ratio to IE$$_2$$ = 1.5)

- IE$$_4$$ = 25024 kJ/mol (ratio to IE$$_3$$ = 6.8) -- Huge jump!

- IE$$_5$$ = 32824 kJ/mol (ratio to IE$$_4$$ = 1.3)

There is a massive jump between IE$$_3$$ (3658) and IE$$_4$$ (25024) -- nearly a 7-fold increase. This indicates that the first three electrons are relatively easy to remove (valence electrons), while the fourth electron comes from an inner, more stable shell.

An element with 3 valence electrons belongs to Group 13 (the Boron group). The electronic configuration would be of the form $$[\text{noble gas}] \, ns^2 \, np^1$$, giving 3 valence electrons before reaching the noble gas core.

The correct answer is Option 1: Group 13.