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Which of the following statements are correct? A. The process of the addition an electron to a neutral gaseous atom is always exothermic B. The process of removing an electron from an isolated gaseous atom is always endothermic C. The 1^textst ionization energy of the boron is less than that of the beryllium D. The electronegativity of C is 2.5 in CH_4 and CCl_4 E. Li is the most electropositive among elements of group I Choose the correct answer from the options gives below

Solution & Explanation

### Core Logic Let us check each criteria statement: * **A is incorrect:** Electron gain can be endothermic for stable configurations like noble gases or alkaline earth metals. * **B is correct:** Removing an electron from a stable atomic nucleus always requires input energy, hence Delta H > 0 (endothermic). * **C is correct:** textBe \ (1s^2 2s^2) has a stable, fully-filled subshell configuration, making its first ionization energy higher than textB \ (1s^2 2s^2 2p^1) where the electron is removed from a higher energy p-orbital. * **D is incorrect:** Due to inductive withdrawal and shifting effective charge distribution, electronegativity alters slightly contextually across different molecular systems (CCl_4 > CH_4). * **E is incorrect:** Cesium (textCs) is the most electropositive Group 1 element. ### Step 1: Match with Choices Statements B and C are definitively evaluated to be correct, corresponding to option (1). ### Pattern Recognition Shortcut: Ionization energy is strictly endothermic (+ Delta H). Beryllium versus Boron is a classic fully-filled subshell anomaly (textIE_1 text Be > textB). Knowing these isolates option (1) immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Reference Study Guides

Classification Of Elements And Periodicity In Properties

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

Q 2025 Electronegativity Trends
Electronic configuration of four elements A, B, C and D are given below : (A) 1mathrms^2 2mathrms^2 2mathrmp^3 (B) 1mathrms^2 2mathrms^2 2mathrmp^4 (C) 1mathrms^2 2mathrms^2 2mathrmp^5 (D) 1mathrms^2 2mathrms^2 2mathrmp^2 Which of the following is the correct order of increasing electronegativity (Pauling's scale)?
  • A. mathrmA < D < B < C
  • B. mathrmA < C < B < D
  • C. mathrmA < B < C < D
  • D. mathrmD < A < B < C

Solution

### Related Formula chi_mathrmP propto Z_texteff propto frac1textAtomic Radius quad text(Across a period) ### Core Logic Let's first identify each element based on its electronic configuration: (A): 1mathrms^2 2mathrms^2 2mathrmp^3 implies textAtomic Number 7 implies textNitrogen (N)

(B): 1mathrms^2 2mathrms^2 2mathrmp^4 implies textAtomic Number 8 implies textOxygen (O)

(C): 1mathrms^2 2mathrms^2 2mathrmp^5 implies textAtomic Number 9 implies textFluorine (F)

(D): 1mathrms^2 2mathrms^2 2mathrmp^2 implies textAtomic Number 6 implies textCarbon (C) ### Step 1: Check Periodic Table Trends All four elements belong to the 2nd period. Electronegativity increases across a period from left to right because nuclear charge (Z_texteff) increases, and atomic radius decreases: textCarbon (D) < textNitrogen (A) < textOxygen (B) < textFluorine (C) On Pauling's scale, the precise values are: - Carbon (D) = 2.55 - Nitrogen (A) = 3.04 - Oxygen (B) = 3.44 - Fluorine (C) = 3.98 ### Step 2: Conclusion The correct increasing order is mathrmD < A < B < C. ### Pattern Recognition Fluorine is the most electronegative element in the entire periodic table (Pauling electronegativity of 4.0). Electronegativity always increases towards the top-right of the main-group elements. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties
Q28 2025 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. text(1) B and D only
  • B. text(2) A and C only
  • C. text(3) B only
  • D. text(4) A, B, C and D

Solution

### Related Formula Standard downward group variations typically follow predictable monotonic pathways: textRadius propto textNumber of shells textIonization Energy propto frac1textAtomic Size ### Core Logic Let's review the explicit values and trends across mathrmF, Cl, Br, I: * **Covalent radius**: mathrmF < Cl < Br < I (Perfect monotonic increase). * **Ionic radius**: mathrmF^- < Cl^- < Br^- < I^- (Perfect monotonic increase). * **First ionization energy**: mathrmF > Cl > Br > I (Perfect monotonic decrease). * **Electron affinity**: mathrmCl > 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 12 Chemistry: The p-Block Elements
Q39 2025 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 mathrmAl^3+ is less than that of mathrmGa^3+. In the light of the above statements, choose the most appropriate answer from the options given below:
  • A. (1)\ textBoth Statement I and Statement II are incorrect
  • B. (2)\ textStatement I is incorrect but Statement II is correct
  • C. (3)\ textStatement I is correct but Statement II is incorrect
  • D. (4)\ textBoth Statement I and Statement II are correct

Solution

### Related Formula Effective nuclear charge scaling expression layout: Z_texteff = Z - sigma ### Core Logic Let's analyze both properties systematically: * **Statement I**: The metallic radius of Gallium (mathrmGa) is anomalously *smaller* than Aluminum (mathrmAl) due to the poor shielding effect of the 10 d-electrons inserted before it. This increases Z_texteff, pulling the outer shell inward tightly. Thus, textRadius of Al > textRadius of Ga, making Statement I incorrect. * **Statement II**: When looking at completely stripped ionic configurations (mathrmAl^3+ and mathrmGa^3+), the extra shell layer in Gallium resumes its dominant role, meaning standard periodic increase down a group holds true: mathrmAl^3+ < Ga^3+. (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+. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 11 Chemistry: The p-Block Elements
Q 2025 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) 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
Q32 2025 Ionization Enthalpy
The atomic number of the element from the following with the lowest 1^textst 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 implies textPotassium (K), an alkali metal in Period 4. * Z = 32 implies textGermanium (Ge), a metalloid in Group 14, Period 4. * Z = 35 implies textBromine (Br), a halogen in Group 17, Period 4. * Z = 87 implies textFrancium (Fr), an alkali metal in Period 7. **Ionization Enthalpy Trends**: 1. Ionization enthalpy increases across a period from left to right due to an increase in effective nuclear charge. 2. Ionization enthalpy decreases down a group due to increasing atomic radius and increasing screening effects. Comparing alkali metals (K and Fr), Francium (Fr, [Rn]7s^1) 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

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