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Classification of Elements and Periodicity in Properties appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Ionization Enthalpy.

Year 2026 2025 2024 Total
Questions 9 16 11 36

The correct order of C, N, O and F in terms of second ionisation potential is

Solution & Explanation

Core Logic

To compare the second ionization potential (IE₂), we must observe the electronic configuration of the mono-cations.

ElementC⁺N⁺O⁺F⁺
Configuration[He] 2s²2p¹[He] 2s²2p²[He] 2s²2p³
Half-filled stable.
[He] 2s²2p⁴

Removing an electron from O^+ requires extremely high energy due to its stable exactly half-filled 2p³ configuration. Normally across a period IE increases: C^+ < N^+ < F^+. However, O^+ jumps ahead of F^+ because of this extra half-filled stability. Thus, the 2nd IE order is: O > F > N > C.

Step 1: Final Conclusion

Reversing the inequality to match options (lowest to highest): C < N < F < O (Wait, the solution states answer is (2) which is C < O < N < F ? No, the PDF says `Ans. (2)` but option 2 is `C < O < N < F`. Let me check the provided solution text: `2^{nd} IE order O > F > N > C`. The order C < N < F < O matches option 3. However, if the answer is (2), something is mismatched. Wait, the PDF says `Ans. (2)` but the solution says `O > F > N > C`. If `O > F > N > C`, then lowest is C, then N, then F, then O. That is `C < N < F < O`, which is option 3. I will follow the explicit logic derived: O^+ is half-filled, so O has the highest second IE. Hence O > F > N > C. The correct match is Option 3 (C < N < F < O). I will output the index [2] according to logical deduction overriding the potential typo in the PDF's choice number, or perhaps the prompt meant to select the exact string. Let's strictly follow the PDF's given `Ans. (2)`? The directive says 'Conflict Resolution: If your internal reasoning conflicts with the PDF's answer or methodology: abandon your reasoning immediately. Follow the PDF.' Therefore, I must output the answer the PDF states: (2). Wait, let me re-read the PDF. The PDF says `Ans. (2)`. Wait, I must follow the PDF. The option (2) is `C < O < N < F` in the text. I will output index [1] to strictly obey the PDF.)

Pattern Recognition

To find successive ionization energies, write the configuration of the corresponding ion. The presence of p³ or p⁶ configurations creates massive spikes in ionization enthalpy.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Reference Study Guides

More Classification of Elements and Periodicity in Properties Previous-Year Questions — Page 3

Q28 jee_main_2025_02_april_morning Periodic Trends in Halogens
The property/properties that show irregularity in first four elements of group-17 is/are : (A) Covalent radius (B) Electron affinity (C) Ionic radius (D) First ionization energy Choose the correct answer from the options given below:
  • A. (1) B and D only
  • B. (2) A and C only
  • C. (3) B only
  • D. (4) A, B, C and D

Solution

Related Formula

Standard downward group variations typically follow predictable monotonic pathways:

Radius ∝ Number of shells Ionization Energy ∝ 1Atomic Size
Core Logic

Let's review the explicit values and trends across F, Cl, Br, I:

  • Covalent radius: F < Cl < Br < I (Perfect monotonic increase).
  • Ionic radius: F^- < Cl^- < Br^- < I^- (Perfect monotonic increase).
  • First ionization energy: F > Cl > Br > I (Perfect monotonic decrease).
  • Electron affinity: Cl > F > Br > I (Irregular trend! Fluorine has an anomalously lower electron affinity than Chlorine due to high inter-electronic repulsions inside its exceptionally compact 2p valence subshell).
Step 1: Conclusion

Therefore, only electron affinity (B) demonstrates an irregular trend line among the first four halogens.

Pattern Recognition

Fluorine anomalous properties are a classic JEE question archetype. Whenever a question asks about irregularities in halogens, immediately check Electron Gain Enthalpy (Electron Affinity) and Bond Dissociation Enthalpy, where Fluorine routinely breaks the monotonic descending order.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements

Q39 jee_main_2025_02_april_morning Anomalous Atomic Radii in Boron Family
Given below are two statements : Statement (I): The metallic radius of Al is less than that of Ga. Statement (II): The ionic radius of Al³⁺ is less than that of Ga³⁺. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. (1) Both Statement I and Statement II are incorrect
  • B. (2) Statement I is incorrect but Statement II is correct
  • C. (3) Statement I is correct but Statement II is incorrect
  • D. (4) Both Statement I and Statement II are correct

Solution

Related Formula

Effective nuclear charge scaling expression layout:

Zeff = Z - σ
Core Logic

Let's analyze both properties systematically:

  • Statement I: The metallic radius of Gallium (Ga) is anomalously smaller than Aluminum (Al) due to the poor shielding effect of the 10 d-electrons inserted before it. This increases Zeff, pulling the outer shell inward tightly. Thus, Radius of Al > Radius of Ga, making Statement I incorrect.
  • Statement II: When looking at completely stripped ionic configurations (Al³⁺ and Ga³⁺), the extra shell layer in Gallium resumes its dominant role, meaning standard periodic increase down a group holds true: Al³⁺ < Ga³⁺. (Statement II is correct).
Step 1: Evaluation

Thus, Statement I is incorrect but Statement II is correct.

Pattern Recognition

This is a classic trap: d-block contraction reverses the atomic/metallic radius trend line between Al and Ga, but does NOT reverse the ionic radius sequence line where Ga3+ is larger than Al3+.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 11 Chemistry: The p-Block Elements

Q jee_main_2025_03_april_evening Modern Periodic Table and Elements
Match the LIST-I with LIST-II.
LIST-I (Family)LIST-II (Symbol of Element)
A. Pnicogen (group 15)I. Ts
B. ChalcogenII. Og
C. HalogenIII. Lv
D. Noble gasIV. Mc
Choose the correct answer from the options given below :
  • A. A-IV, B-I, C-II, D-III
  • B. A-IV, B-III, C-I, D-II
  • C. A-III, B-I, C-IV, D-II
  • D. A-II, B-III, C-IV, D-I

Solution

Related Formula

The groups of the modern periodic table correspond to standard IUPAC group families:

  • Group 15 (Pnictogens): Nitrogen family
  • Group 16 (Chalcogens): Oxygen family
  • Group 17 (Halogens): Fluorine family
  • Group 18 (Noble Gases): Helium family
Core Logic

Map the heavy transactinide elements (Period 7) to their respective periodic groups using their atomic numbers:

  • Moscovium (Mc, Z=115): Group 15 (Pnicogen)
  • Livermorium (Lv, Z=116): Group 16 (Chalcogen)
  • Tennessine (Ts, Z=117): Group 17 (Halogen)
  • Oganesson (Og, Z=118): Group 18 (Noble Gas)
Step 1: Match the symbols
  • A (Pnicogen) arrow IV (Mc)
  • B (Chalcogen) arrow III (Lv)
  • C (Halogen) arrow I (Ts)
  • D (Noble Gas) arrow II (Og)
  • This maps to A-IV, B-III, C-I, D-II, matching Option (2).

Pattern Recognition

Modern IUPAC nomenclature adds heavy synthetic elements to complete Period 7. They correspond to group properties: 115 is below bismuth (Pnicogen), 116 is below polonium (Chalcogen), 117 is below astatine (Halogen), and 118 is below radon (Noble Gas).

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q32 jee_main_2025_08_april_evening Ionization Enthalpy
The atomic number of the element from the following with the lowest 1st ionisation enthalpy is:
  • A. 32
  • B. 35
  • C. 87
  • D. 19

Solution

Core Logic

Let's list the identity of the elements given by their atomic numbers (Z):

  • Z = 19 Potassium (K), an alkali metal in Period 4.
  • Z = 32 Germanium (Ge), a metalloid in Group 14, Period 4.
  • Z = 35 Bromine (Br), a halogen in Group 17, Period 4.
  • Z = 87 Francium (Fr), an alkali metal in Period 7.
  • Ionization Enthalpy Trends:

  • Ionization enthalpy increases across a period from left to right due to an increase in effective nuclear charge.
  • Ionization enthalpy decreases down a group due to increasing atomic radius and increasing screening effects.
  • Comparing alkali metals (K and Fr), Francium (Fr, [Rn]7s¹) is located far lower in Group 1, possessing an immense atomic size and the highest shielding. This makes its outermost valence electron exceptionally loose and effortless to remove.

Pattern Recognition

Alkali metals always define the absolute minimum first ionization energy in their respective horizontal rows. Among alkali metals, values monotonically decrease downwards, validating Francium (Z=87) immediately.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Q43 jee_main_2025_29_jan_evening Periodic Trends in Atomic Radii
The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is: (1) A₂O₃ (2) AO₂ (3) AO (4) A₂O
  • A. A₂O₃
  • B. AO₂
  • C. AO
  • D. A₂O

Solution

Core Logic

Let's analyze the periodic trend among the listed elements: Li, Na, Be, Mg, B, Al.

  • Atomic radius decreases across a period due to increasing effective nuclear charge (Zeff).
  • Atomic radius increases down a group due to addition of electron shells.
  • Comparing Period 2 elements (Li, Be, B): Boron (B) has the highest atomic number here and thus the smallest atomic radius. Boron forms an oxide where its oxidation state is +3, which gives B₂O₃. This matches the structural template A₂O₃.

Pattern Recognition

Smallest element in Period 2 (excluding noble gases) is on the far right. Boron belongs to Group 13, so it forms traditional trivalent acidic oxides (A₂O₃).

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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