### Related Formula chi_mathrmP propto Z_texteff propto frac1textAtomic Radius quad text(Across a period)$\chi_{\mathrm{P}} \propto Z_{\text{eff}} \propto \frac{1}{\text{Atomic Radius}} \quad \text{(Across a period)}$ ### 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)$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^3 \implies \text{Atomic Number } 7 \implies \text{Nitrogen (N)}$

(B): 1mathrms^2 2mathrms^2 2mathrmp^4 implies textAtomic Number 8 implies textOxygen (O)$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^4 \implies \text{Atomic Number } 8 \implies \text{Oxygen (O)}$

(C): 1mathrms^2 2mathrms^2 2mathrmp^5 implies textAtomic Number 9 implies textFluorine (F)$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^5 \implies \text{Atomic Number } 9 \implies \text{Fluorine (F)}$

(D): 1mathrms^2 2mathrms^2 2mathrmp^2 implies textAtomic Number 6 implies textCarbon (C)$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^2 \implies \text{Atomic Number } 6 \implies \text{Carbon (C)}$ ### 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$Z_{\text{eff}}$) increases, and atomic radius decreases: textCarbon (D) < textNitrogen (A) < textOxygen (B) < textFluorine (C)$\text{Carbon (D)} < \text{Nitrogen (A)} < \text{Oxygen (B)} < \text{Fluorine (C)}$ On Pauling's scale, the precise values are: - Carbon (D) = 2.55$2.55$ - Nitrogen (A) = 3.04$3.04$ - Oxygen (B) = 3.44$3.44$ - Fluorine (C) = 3.98$3.98$ ### Step 2: Conclusion The correct increasing order is mathrmD < A < B < C$\mathrm{D < A < B < C}$. ### Pattern Recognition Fluorine is the most electronegative element in the entire periodic table (Pauling electronegativity of 4.0$4.0$). 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

### Related Formula Standard downward group variations typically follow predictable monotonic pathways: textRadius propto textNumber of shells$\text{Radius} \propto \text{Number of shells}$ textIonization Energy propto frac1textAtomic Size$\text{Ionization Energy} \propto \frac{1}{\text{Atomic Size}}$ ### Core Logic Let's review the explicit values and trends across mathrmF, Cl, Br, I$\mathrm{F, Cl, Br, I}$: * **Covalent radius**: mathrmF < Cl < Br < I$\mathrm{F < Cl < Br < I}$ (Perfect monotonic increase). * **Ionic radius**: mathrmF^- < Cl^- < Br^- < I^-$\mathrm{F^- < Cl^- < Br^- < I^-}$ (Perfect monotonic increase). * **First ionization energy**: mathrmF > Cl > Br > I$\mathrm{F > Cl > Br > I}$ (Perfect monotonic decrease). * **Electron affinity**: mathrmCl > F > Br > I$\mathrm{Cl > F > Br > I}$ (Irregular trend! Fluorine has an anomalously lower electron affinity than Chlorine due to high inter-electronic repulsions inside its exceptionally compact 2p$2p$ 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

### Related Formula Effective nuclear charge scaling expression layout: Z_texteff = Z - sigma$Z_{\text{eff}} = Z - \sigma$ ### Core Logic Let's analyze both properties systematically: * **Statement I**: The metallic radius of Gallium (mathrmGa$\mathrm{Ga}$) is anomalously *smaller* than Aluminum (mathrmAl$\mathrm{Al}$) due to the poor shielding effect of the 10$10$ d-electrons inserted before it. This increases Z_texteff$Z_{\text{eff}}$, pulling the outer shell inward tightly. Thus, textRadius of Al > textRadius of Ga$\text{Radius of Al} > \text{Radius of Ga}$, making Statement I incorrect. * **Statement II**: When looking at completely stripped ionic configurations (mathrmAl^3+$\mathrm{Al^{3+}}$ and mathrmGa^3+$\mathrm{Ga^{3+}}$), the extra shell layer in Gallium resumes its dominant role, meaning standard periodic increase down a group holds true: mathrmAl^3+ < Ga^3+$\mathrm{Al^{3+} < Ga^{3+}}$. (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

### 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$Z=115$): Group 15 (Pnicogen) - Livermorium (Lv, Z=116$Z=116$): Group 16 (Chalcogen) - Tennessine (Ts, Z=117$Z=117$): Group 17 (Halogen) - Oganesson (Og, Z=118$Z=118$): Group 18 (Noble Gas) ### Step 1: Match the symbols - A (Pnicogen) rightarrow$\rightarrow$ IV (Mc) - B (Chalcogen) rightarrow$\rightarrow$ III (Lv) - C (Halogen) rightarrow$\rightarrow$ I (Ts) - D (Noble Gas) rightarrow$\rightarrow$ 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$115$ is below bismuth (Pnicogen), 116$116$ is below polonium (Chalcogen), 117$117$ is below astatine (Halogen), and 118$118$ is below radon (Noble Gas). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic Let's list the identity of the elements given by their atomic numbers (Z$Z$): * Z = 19 implies textPotassium (K)$Z = 19 \implies \text{Potassium (K)}$, an alkali metal in Period 4. * Z = 32 implies textGermanium (Ge)$Z = 32 \implies \text{Germanium (Ge)}$, a metalloid in Group 14, Period 4. * Z = 35 implies textBromine (Br)$Z = 35 \implies \text{Bromine (Br)}$, a halogen in Group 17, Period 4. * Z = 87 implies textFrancium (Fr)$Z = 87 \implies \text{Francium (Fr)}$, 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$K$ and Fr$Fr$), Francium (Fr, [Rn]7s^1$Fr, [Rn]7s^1$) 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$Z=87$) immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties