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 :

Solution & Explanation

### 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) rightarrow IV (Mc) - B (Chalcogen) rightarrow III (Lv) - C (Halogen) rightarrow I (Ts) - D (Noble Gas) rightarrow 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). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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

Q52 jee_main_2026_21_jan_morning Periodic Trends
Which of the following represents the correct trend for the mentioned property? A. F > P > S > B – First Ionization Energy B. Cl > F > S > P – Electron Affinity C. K > Al > Mg > B – Metallic character D. K_2O > Na_2O > MgO > Al_2O_3 – Basic character Choose the correct answer from the option given below.
  • A. textA, B and D only
  • B. textA, B, C and D
  • C. textA and B only
  • D. textB and C only

Solution

### Core Logic Analyzing each statement based on periodic trends: A. On moving left to right in a period, Ionization Energy (IE) generally increases, and from top to bottom it decreases. So, the correct order is F > P > S > B (IE order). Thus, statement A is correct. B. For Electron Affinity (EA), Group 17 > Group 16 > Group 15. Also, 3rd-period elements often have higher EA than 2nd period (like Cl > F due to compact size of F). The order Cl > F > S > P is correct. Thus, statement B is correct. C. On moving left to right in a period, metallic character decreases. So Mg > Al. The correct order is K > Mg > Al > B. Thus, statement C is incorrect. D. On moving top to bottom in a group basic character increases, and moving left to right it decreases. The correct basic strength order is K_2O > Na_2O > MgO > Al_2O_3. Thus, statement D is correct. ### Step 1: Conclusion Statements A, B, and D represent the correct trends. ### Pattern Recognition Always remember the electron affinity anomaly: Cl > F and S > O due to high inter-electronic repulsion in smaller 2p orbitals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q60 jee_main_2026_21_jan_evening Atomic/Ionic Radii and Electron Gain Enthalpy
Given below are two statements: Statement-I: The correct order in terms of atomic/ionic radii is textAl > textMg > textMg^2+ > textAl^3+. Statement-II: The correct order in terms of the magnitude of electron gain enthalpy is textCl > textBr > textS > textO. In the light of the above statements, choose the correct answer from the options given below:
  • A. (1) \ textBoth Statement I and Statement II are false
  • B. (2) \ textStatement I is false but Statement II is true
  • C. (3) \ textStatement I is true but Statement II is false
  • D. (4) \ textBoth Statement I and Statement II are true

Solution

### Core Logic - Statement I: Correct order of size is textMg > textAl > textMg^2+ > textAl^3+ because atomic radius of magnesium is greater than aluminium in period 3. Thus Statement-I is false. - Statement-II: Chlorine has the highest electron gain enthalpy in the periodic table, and halogens exceed chalcogens. The order textCl > textBr > textS > textO is true. ### Step 1: Final Conclusion Statement I is false but Statement II is true, corresponding to option (2). ### Pattern Recognition Sees: Periodic trends for atomic radii and electron affinity. Trap: Assuming Al is larger than Mg due to higher atomic number. ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q62 jee_main_2026_22_january_evening Ionization Enthalpy and Electron Gain Enthalpy Trends
Given below are two statements: Statement-I: textC < textO < textN < textF is the correct order in terms of first ionization enthalpy values. Statement-II: textS > textSe > textTe > textPo > textO is the correct order in terms of the magnitude of electron gain enthalpy values. 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. Both Statement-I and Statement-II are false.
  • D. Statement-I is true but Statement-II is false.

Solution

### Related Formula textHalf-filled 2p^3 text configuration of Nitrogen gives higher IE_1 text than Oxygen (2p^4text). textOxygen has anomalously low magnitude of Delta_egH text due to strong inter-electronic repulsions in small 2p text shell. ### Core Logic Step 1: Evaluate Statement-I: - Across Period 2, IE_1 generally increases with Z_texteff. - N (2p^3) is half-filled, so IE_1(textN) > IE_1(textO). - Correct order: textC < textO < textN < textF. Statement-I is TRUE. Step 2: Evaluate Statement-II: - Magnitudes of Delta_egH for Group 16: textS (200) > textSe (195) > textTe (190) > textPo (174) > textO (141text kJ/mol). - Oxygen has the lowest magnitude in the group. Statement-II is TRUE. ### Pattern Recognition Sees: Group 16 electron gain enthalpy and Period 2 ionization enthalpy anomalies. Shortcut: Remember half-filled N > O for IE_1, and small 2p shell makes O < Po for |Delta_egH|. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q53 jee_main_2026_23_january_morning Ionization Enthalpy
The correct trend in the first ionization enthalpies of the elements in the 3^textrd period of periodic table is:
  • A. Al < Si < S < P < Cl
  • B. Al < S < P < Si < Cl
  • C. Si < S < Al < P < Cl
  • D. S < Si < Al < P < Cl

Solution

### Core Logic In general, on moving from left to right across a period, the first ionization energy increases due to an increase in effective nuclear charge (Z_texteff). However, there are exceptions due to stable electronic configurations. ### Step 1: Configuration Analysis For elements Al, Si, P, S, and Cl: Generally, Al < Si < P < S < Cl. But, Phosphorus (1s^2 2s^2 2p^6 3s^2 3p^3) has a half-filled, exceptionally stable 3p subshell compared to Sulfur (3s^2 3p^4). This makes it harder to remove an electron from P than from S. ### Step 2: Final Trend Construction Because of this half-filled stability, the ionization energy of P is greater than that of S. Therefore, the corrected trend becomes: Al < Si < S < P < Cl ### Pattern Recognition Always look for Group 15 (half-filled np^3) vs Group 16 (np^4) anomalies. Group 15 always has a higher first ionization energy than Group 16 in the same period. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q59 jee_main_2026_23_january_evening Ionization Enthalpy and Ionic Radius
Ionization Enthalpy and Ionic Radius diagram for Q59 - JEE Main 2026 Evening
Relevant data for evaluating Statement I and II regarding atomic properties.
Given below are two statements : Statement I : The second ionisation enthalpy of Na is larger than the corresponding ionisation enthalpy of Mg. Statement II : The ionic radius of O^2- is larger than that of F^-. In the light of the above statements, choose the correct answer from the options given below.
  • A. textBoth statement I and statement II are true
  • B. textBoth statement I and statement II are false
  • C. textStatement I is false but statement II is true
  • D. textStatement I is true but statement II is false

Solution

### Related Formula IE_2 text requires breaking stable noble gas configurations if M^+ text is isoelectronic with a noble gas. ### Core Logic Statement I: Let's analyze the electronic configurations. Na (Z=11): 1s^2 2s^2 2p^6 3s^1 implies Na^+ is 1s^2 2s^2 2p^6 (Stable Neon noble gas core). Mg (Z=12): 1s^2 2s^2 2p^6 3s^2 implies Mg^+ is 1s^2 2s^2 2p^6 3s^1. Removing a second electron from Na^+ (IE_2) involves disrupting a highly stable, fully-filled 2p^6 shell, requiring massive energy. Removing a second electron from Mg^+ (IE_2) just removes the 3s^1 electron. Thus, IE_2 of Na > IE_2 of Mg. Statement I is true. Statement II: Both O^2- and F^- are isoelectronic species, possessing 10 electrons (1s^2 2s^2 2p^6). However, the nuclear charge (number of protons, Z) is different. O^2- has 8 protons pulling 10 electrons. F^- has 9 protons pulling 10 electrons. Since F^- has a higher effective nuclear charge (Z_texteff), its electron cloud is pulled more tightly, making its radius smaller. Thus, the radius of O^2- > F^-. Statement II is true. ### Pattern Recognition For isoelectronic species, more negative charge always equals a larger ionic radius (lower Z/e ratio implies less nuclear pull per electron). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity
Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)