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 :

Correct answer: A. Group 13 The successive 5 ionisation energies of an element are 800,2427,3658,25024 and 32824 ~kJ / mol , respectively. aligned & IE_2 IE_1 =(2427)/(800)=3.03 & IE IE_2 =(3658)/(2427)=1.51 & IE IE_3 =(25024)/(3658)=6.84 aligned IE_5 IE_4 =(32824)/(25024)=1.31 Since ( IE_4 / IE_3 ) value is maximum, the element belongs to group 13.

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Classification of Elements and Periodicity in Properties

JEE Main 2025 · 24 January · Evening Shift · Chemistry · Classification of Elements and Periodicity in Properties

The successive 5 ionisation energies of an element are...

The successive 5 ionisation energies of an element are

800,2427,3658,25024

and

32824 \mathrm{~kJ} / \mathrm{mol}

, respectively. By using the above values predict the group in which the above element is present :

Options

AGroup 13
BGroup 14
CGroup 2
DGroup 4

Correct answer

A. Group 13

Step-by-step solution

The successive 5 ionisation energies of an element are 800,2427,3658,25024 and

32824 \mathrm{~kJ} / \mathrm{mol}

, respectively.

\begin{aligned} & \frac{\mathrm{IE}_2}{\mathrm{IE}_1}=\frac{2427}{800}=3.03 \\ & \frac{\mathrm{IE}}{\mathrm{IE}_2}=\frac{3658}{2427}=1.51 \\ & \frac{\mathrm{IE}}{\mathrm{IE}_3}=\frac{25024}{3658}=6.84 \end{aligned}

\frac{\mathrm{IE}_5}{\mathrm{IE}_4}=\frac{32824}{25024}=1.31

Since (

\mathrm{IE}_4 / \mathrm{IE}_3$ ) value is maximum, the element belongs to group 13.

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