Given below are two statements: Statement-I: textC < textO < textN < textF$\text{C} < \text{O} < \text{N} < \text{F}$ is the correct order in terms of first ionization enthalpy values. Statement-II: textS > textSe > textTe > textPo > textO$\text{S} > \text{Se} > \text{Te} > \text{Po} > \text{O}$ 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:

### 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

### Core Logic - Statement I: Correct order of size is textMg > textAl > textMg^2+ > textAl^3+$\text{Mg} > \text{Al} > \text{Mg}^{2+} > \text{Al}^{3+}$ 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$\text{Cl} > \text{Br} > \text{S} > \text{O}$ 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

### 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