Which of the following represents the correct trend for the mentioned property? A. F > P > S > B$F > P > S > B$ – First Ionization Energy B. Cl > F > S > P$Cl > F > S > P$ – Electron Affinity C. K > Al > Mg > B$K > Al > Mg > B$ – Metallic character D. K_2O > Na_2O > MgO > Al_2O_3$K_{2}O > Na_{2}O > MgO > Al_{2}O_{3}$ – Basic character Choose the correct answer from the option 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