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:

Let's examine the differences between successive ionization energies to find where the largest jump occurs: * IE1 = 800 kJ/mol * IE2 = 2427 kJ/mol * IE3 = 3658 kJ/mol * IE4 = 25024 kJ/mol * IE5 = 32824 kJ/mol Notice the massive, multi-fold jump between IE3 and IE4 (3658 arrow 25024 kJ/mol). This huge increase indicates that removing the 4th electron requires breaking into a stable, filled core noble gas shell configuration. This means the atom has exactly 3 valence electrons in its outermost shell (ns2 np1), which identifies it as a member of **Group 13** (the Boron family). To find the number of valence electrons, look for the ionization step where the value jumps drastically. The number of relatively low ionization steps before that jump equals the number of valence electrons. Here, 3 low steps arrow 3 valence electrons ightarrow Group 13. Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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:

Let's examine the differences between successive ionization energies to find where the largest jump occurs: * IE1 = 800 kJ/mol * IE2 = 2427 kJ/mol * IE3 = 3658 kJ/mol * IE4 = 25024 kJ/mol * IE5 = 32824 kJ/mol Notice the massive, multi-fold jump between IE3 and IE4 (3658 arrow 25024 kJ/mol). This huge increase indicates that removing the 4th electron requires breaking into a stable, filled core noble gas shell configuration. This means the atom has exactly 3 valence electrons in its outermost she

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:

Let's examine the differences between successive ionization energies to find where the largest jump occurs: * IE1 = 800 kJ/mol * IE2 = 2427 kJ/mol * IE3 = 3658 kJ/mol * IE4 = 25024 kJ/mol * IE5 = 32824 kJ/mol Notice the massive, multi-fold jump between IE3 and IE4 (3658 arrow 25024 kJ/mol). This huge increase indicates that removing the 4th electron requires breaking into a stable, filled core noble gas shell configuration. This means the atom has exactly 3 valence electrons in its outermost shell (ns2 np1), which identifies it as a member of **Group 13** (the Boron family). To find the number of valence electrons, look for the ionization step where the value jumps drastically. The number of relatively low ionization steps before that jump equals the number of valence electrons. Here, 3 low steps arrow 3 valence electrons ightarrow Group 13. Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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:

Let's examine the differences between successive ionization energies to find where the largest jump occurs: * IE1 = 800 kJ/mol * IE2 = 2427 kJ/mol * IE3 = 3658 kJ/mol * IE4 = 25024 kJ/mol * IE5 = 32824 kJ/mol Notice the massive, multi-fold jump between IE3 and IE4 (3658 arrow 25024 kJ/mol). This huge increase indicates that removing the 4th electron requires breaking into a stable, filled core noble gas shell configuration. This means the atom has exactly 3 valence electrons in its outermost she

The successive 5 ionisation energies of an element are 800, 2427, 3658, 25024 and 32824 kJ/mol, respectively. By using…

Year	2026	2025	2024	Total
Questions	9	16	11	36

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Let's examine the differences between successive ionization energies to find where the largest jump occurs:

$IE₁ = 800 kJ/mol$ $\text{IE}_1 = 800\text{ kJ/mol}$
$IE₂ = 2427 kJ/mol$ $\text{IE}_2 = 2427\text{ kJ/mol}$
$IE₃ = 3658 kJ/mol$ $\text{IE}_3 = 3658\text{ kJ/mol}$
$IE₄ = 25024 kJ/mol$ $\text{IE}_4 = 25024\text{ kJ/mol}$
$IE₅ = 32824 kJ/mol$ $\text{IE}_5 = 32824\text{ kJ/mol}$

Notice the massive, multi-fold jump between $IE₃$ $\text{IE}_3$ and $IE₄$ $\text{IE}_4$ ( $3658 arrow 25024 kJ/mol$ $3658 \rightarrow 25024\text{ kJ/mol}$).

This huge increase indicates that removing the 4th electron requires breaking into a stable, filled core noble gas shell configuration. This means the atom has exactly $3$ $3$ valence electrons in its outermost shell ( $ns² np¹$ $ns^2 np^1$), which identifies it as a member of Group 13 (the Boron family).

Pattern Recognition

To find the number of valence electrons, look for the ionization step where the value jumps drastically. The number of relatively low ionization steps before that jump equals the number of valence electrons. Here, 3 low steps $arrow$ $\rightarrow$ 3 valence electrons $arrow$ $\rightarrow$ Group 13.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 8

Q70 jee_main_2024_31_jan_morning Electron Gain Enthalpy

The correct sequence of electron gain enthalpy of the elements listed below is A. Ar B. Br C. F D. S Choose the most appropriate from the options given below:

A. $C > B > D > A$ $C > B > D > A$
B. $A > D > B > C$ $A > D > B > C$
C. $A > D > C > B$ $A > D > C > B$
D. $D > C > B > A$ $D > C > B > A$

Solution

Core Logic

Electron gain enthalpy ( $ΔegH$ $\Delta_{eg}H$) values are considered with their signs when ordering from largest to smallest algebraically.

Element	$ΔegH (kJ/mol)$ $\Delta_{eg}H \text{ (kJ/mol)}$
F	-333
S	-200
Br	-325
Ar	+96

Arranging these algebraically from highest to lowest: $+96 > -200 > -325 > -333$ $+96 > -200 > -325 > -333$ Thus, $Ar > S > Br > F$ $Ar > S > Br > F$, which corresponds to $A > D > B > C$ $A > D > B > C$.

Pattern Recognition

Noble gases always have positive electron gain enthalpies (highest algebraically). Halogens have highly negative values (lowest algebraically). Fluorine is highly negative, Bromine is slightly less negative. Sulfur sits in between.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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 & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 8

Solution

Core Logic

Pattern Recognition

Chapter Mix

More Classification of Elements and Periodicity in Properties Questions — jee_main_2025_24_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

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 & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 8

Solution

Core Logic

Pattern Recognition

Chapter Mix

More Classification of Elements and Periodicity in Properties Questions — jee_main_2025_24_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back