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…

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More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 5

Q35 jee_main_2025_24_jan_morning Modern Periodic Law and Periodic Table Structure

Which of the following Statements are NOT true about the periodic table? A. The properties of elements are function of atomic weights. B. The properties of elements are function of atomic numbers. C. Elements having similar outer electronic configuration are arranged in same period. D. An element's location reflects the quantum numbers of the last filled orbital. E. The number of elements in a period is same as the number of atomic orbitals available in energy level that is being filled. Choose the correct answer from the options given below:

A. A, C and E Only
B. D and E Only
C. A and E Only
D. B, C and E Only

Solution

Core Logic

Evaluating modern electronic system structures:

Statement A is false: Modern periodic organization links traits to atomic numbers, not weights (which defined old Mendeleev frameworks).
Statement C is false: Elements possessing identical outer configurations occupy the same chemical group, not the same horizontal period.
Statement E is false: The total count of active elements across any discrete period is equal to double ( $2 \times$ $2 \times$) the count of functional available atomic orbitals in the relevant energy level.

Pattern Recognition

Carefully identify double-negatives or questions asking for the 'NOT true' choice inside variable selection items.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q39 jee_main_2025_28_jan_evening Genesis of Periodic Classification

Given below are two statements: Statement (I): According to the Law of Octaves, the elements were arranged in the increasing order of their atomic number. Statement (II): Meyer observed a periodically repeated pattern upon plotting physical properties of certain elements against their respective atomic numbers. In the light of the above statements, choose the correct answer from the options given below:

A. Statement I is false but Statement II is true
B. Both Statement I and Statement II are true
C. Statement I is true but Statement II is false
D. Both Statement I and Statement II are false

Solution

Related Formula

Early historical periodic classification principles depended on atomic weight attributes:

Periodic Property = f(Atomic Weight)

Core Logic

Evaluating historical accuracy:

Statement I: Newlands' Law of Octaves arranged elements in the increasing order of their atomic weights (not atomic numbers). Hence, Statement I is false.
Statement II: Lothar Meyer plotted physical properties like atomic volume, melting point, and boiling point against atomic weight (not atomic number). Hence, Statement II is false.

Step 1: Conclusion

Since both statements incorrectly reference atomic number instead of atomic weight, both Statement I and Statement II are false.

Pattern Recognition

Almost all classical periodic classifiers (Newlands, Döbereiner, de Chancourtois, Lothar Meyer, Mendeleev) relied strictly on atomic weight. The pivot to atomic number occurred later with Moseley's X-ray studies and the Modern Periodic Law.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q jee_main_2025_29_jan_morning Periodic Trends in Properties

An element 'E' has the ionisation enthalpy value of

374 kJ mol⁻¹

. 'E' reacts with elements A, B, C and D with electron gain enthalpy values of

-328

-349

-325

and

-295 kJ mol⁻¹

, respectively. The correct order of the products EA, EB, EC and ED in terms of ionic character is :

A. $EB > EA > EC > ED$
B. $ED > EC > EA > EB$
C. $EA > EB > EC > ED$
D. $ED > EC > EB > EA$

Solution

Related Formula

Ionic Character \propto Δ HIE - Δ HEGE

Core Logic

The relative ionic quality of a standard binary bond rises as the gap scale between ionization enthalpy and negative electron gain enthalpy parameters widens. Comparing the values:

For B: $Δ HEGE = -349 ~kJ/mol$ (largest energy release) $arrow$ $\rightarrow$ Highest ionic character.
For A: $Δ HEGE = -328 ~kJ/mol$ .
For C: $Δ HEGE = -325 ~kJ/mol$ .
For D: $Δ HEGE = -295 ~kJ/mol$ (smallest energy release) $arrow$ $\rightarrow$ Lowest ionic character.

Arranging them in descending order of ionic character yields:

EB gt EA gt EC gt ED

Pattern Recognition

A highly exothermic electron gain enthalpy value favors easier anion production, widening electronegativity variations to enhance ionic bond properties.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q jee_main_2025_29_jan_morning Periodic Trends in Physical and Chemical Properties

Given below are two statements : Statement (I) : The radii of isoelectronic species increases in the order:

M g ^ 2 + < N a ^ + < F ^ - < O ^ 2 -

Statement (II) : The magnitude of electron gain enthalpy of halogen decreases in the order:

C l > F > B r > I

A. Statement I is incorrect but Statement II is correct
B. Both Statement I and Statement II are incorrect.
C. Statement I is correct but Statement II is incorrect
D. Both Statement I and Statement II are correct

Solution

Related Formula

Ionic Radius \propto 1Nuclear Charge (Z) (for Isoelectronic series)

Core Logic

Evaluating each statement systematically :

Statement (I) is correct: $Mg²⁺, Na⁺, F⁻, O²⁻$ all possess exactly 10 electrons (isoelectronic). As the positive nuclear charge decreases ( $Z = 12$ $Z = 12$ for $Mg$ $\mathrm{Mg}$ down to $Z = 8$ $Z = 8$ for $O$ $\mathrm{O}$), the nucleus exerts less pull on the electron cloud, causing the ionic radius to increase :

Mg²⁺ < Na⁺ < F⁻ < O²⁻

Statement (II) is correct: Chlorine has a higher electron gain enthalpy magnitude than fluorine due to lower electron-electron repulsion in its larger $3p$ $3p$ orbital. The standard halogen trend follows:

Cl > F > Br > I

Thus, both statements are correct.

Pattern Recognition

For species with the same number of electrons, a higher negative charge always leads to a larger electron cloud radius due to reduced nuclear traction.

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 5

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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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 5

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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