Match the LIST-I with LIST-II.

LIST-I (Family)	LIST-II (Symbol of Element)
A. Pnicogen (group 15)	I. Ts
B. Chalcogen	II. Og
C. Halogen	III. Lv
D. Noble gas	IV. Mc

Choose the correct answer from the options given below :

### 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$F > P > S > B$ (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$Cl > F > S > P$ is correct. Thus, statement B is correct. C. On moving left to right in a period, metallic character decreases. So Mg > Al$Mg > Al$. The correct order is K > Mg > Al > B$K > Mg > Al > B$. 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$K_{2}O > Na_{2}O > MgO > Al_{2}O_{3}$. 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$Cl > F$ and S > O$S > O$ 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

### 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