The successive 5 ionisation energies of an element are 800,2427,3658,25024 and 32824\; kJ\,mol -1 , respectively. By using the above values predict the group in which the above element is present :

Chemistry · Periodicity of Elements and Periodic Properties

JEE Main 2025 — 24 January, Evening Shift — Question 41

The successive 5 ionisation energies of an element are $800,2427,3658,25024$ and $32824\; \mathrm{kJ\,mol^{-1}}$ , respectively. By using the above values predict the group in which the above element is present :

Option A:
Group 2
Option B:
Group 13
Correct
Option C:
Group 4
Option D:
Group 14

Answer: B

Step-by-step solution

Successive ionisation energies, we have

$IE_1 = 800$

$IE_2 = 2427$

$IE_3 = 3658$

$IE_4 = 25024$

There is a very large increase between $IE_3$ and $IE_4$ .

This indicates that after removal of three electrons, the next electron is removed from an inner shell.

Hence, the element has three valence electrons.

Therefore, the element belongs to Group $13$ .

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Chemistry
Chapter: Periodicity of Elements and Periodic Properties
Topic: Periodicity in Chemical Properties (Ionization Energy, Electron Affinity, Electronegativity)

Given below are two statements, one is labelled as Statement I and the other is labelled as Statement II. Statement I : The first…

Answer: B

Step-by-step solution

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