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 :

Solution

Chemistry ma’am · 3 steps

Look for the one huge jump in the list. Every electron removed before it is a valence electron.

1. Why jumps appear. Each successive ionisation energy rises a little, because the ion left behind is more positive. A very large jump comes only when you start pulling electrons from a new, inner shell that is much closer to the nucleus.

   Step	IE (kJ mol⁻¹)	Ratio to previous IE
   IE₁	800	-
   IE₂	2427	about 3.0
   IE₃	3658	about 1.5
   IE₄	25024	about 6.8
   IE₅	32824	about 1.3

2. Find the shell break. The jump from IE₃ to IE₄ (3658 to 25024) is far bigger than any other. So the first three electrons come from the outer shell, and the fourth has to come from the inner core.
3. So. The element has 3 valence electrons (ns²np¹), which places it in Group 13. These values belong to boron. The answer is (c).

The idea to keep. In successive ionisation energies, the number of electrons removed before the biggest jump equals the number of valence electrons. That count gives the group: 1, 2, 13, 14 and so on.

Answer from the NTA final answer key. Solution by PrepWiser.