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 :

Periodic Table and Periodicity PYQ — JEE Main 2025 · 24 Jan · Shift 2 · Q1 with Solution

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Correct answer: C

Use the pattern in successive ionisation energies

The given ionisation energies are:
$\quad IE_2 = 2427, \quad IE_3 = 3658, \quad IE_4 = 25024, \quad IE_5 = 32824\ \text{kJ/mol}$$$
Look for the biggest jump

Compute the increases:
- From $IE_1$ to $IE_2$ : $2427 - 800 = 1627$
- From $IE_2$ to $IE_3$ : $3658 - 2427 = 1231$
- From $IE_3$ to $IE_4$ : $25024 - 3658 = 21366$
- From $IE_4$ to $IE_5$ : $32824 - 25024 = 7800$
The largest jump is between $IE_3$ and $IE_4$ .
Interpretation of the large jump

A very large jump after removing 3 electrons means that the atom has 3 valence electrons.

After the first 3 electrons are removed, the next electron must be removed from an inner shell, which requires much more energy.
Relate valence electrons to group

An element with 3 valence electrons belongs to Group 13.
Check options
- A: Group 4 $\rightarrow$ would suggest 4 valence electrons, not correct
- B: Group 2 $\rightarrow$ would show a large jump after $IE_2$ , not correct
- C: Group 13 $\rightarrow$ large jump after $IE_3$ , correct
- D: Group 14 $\rightarrow$ would show a large jump after $IE_4$ , not correct

Therefore, the element belongs to Group 13.