Given below are two statements : Statement I : textK > textMg > textAl > textB$\text{K} > \text{Mg} > \text{Al} > \text{B}$ is the correct order in terms of metallic character. Statement II : Atomic radius is always greater than the ionic radius for any element. 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 textHalf-filled 2p^3 text configuration of Nitrogen gives higher IE_1 text than Oxygen (2p^4text).$$\text{Half-filled } 2p^3 \text{ configuration of Nitrogen gives higher } IE_1 \text{ than Oxygen (}2p^4\text{).}$$ textOxygen has anomalously low magnitude of Delta_egH text due to strong inter-electronic repulsions in small 2p text shell.$$\text{Oxygen has anomalously low magnitude of } \Delta_{eg}H \text{ due to strong inter-electronic repulsions in small } 2p \text{ shell.}$$ ### Core Logic Step 1: Evaluate Statement-I: - Across Period 2, IE_1$IE_1$ generally increases with Z_texteff$Z_{\text{eff}}$. - N (2p^3$2p^3$) is half-filled, so IE_1(textN) > IE_1(textO)$IE_1(\text{N}) > IE_1(\text{O})$. - Correct order: textC < textO < textN < textF$\text{C} < \text{O} < \text{N} < \text{F}$. Statement-I is TRUE. Step 2: Evaluate Statement-II: - Magnitudes of Delta_egH$\Delta_{eg}H$ for Group 16: textS (200) > textSe (195) > textTe (190) > textPo (174) > textO (141text kJ/mol)$\text{S} (200) > \text{Se} (195) > \text{Te} (190) > \text{Po} (174) > \text{O} (141\text{ kJ/mol})$. - Oxygen has the lowest magnitude in the group. Statement-II is TRUE. ### Pattern Recognition Sees: Group 16 electron gain enthalpy and Period 2 ionization enthalpy anomalies. Shortcut: Remember half-filled N > O for IE_1$IE_1$, and small 2p shell makes O < Po for |Delta_egH|$|\Delta_{eg}H|$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic In general, on moving from left to right across a period, the first ionization energy increases due to an increase in effective nuclear charge (Z_texteff$Z_{\text{eff}}$). However, there are exceptions due to stable electronic configurations. ### Step 1: Configuration Analysis For elements Al, Si, P, S, and Cl: Generally, Al < Si < P < S < Cl$Al < Si < P < S < Cl$. But, Phosphorus (1s^2 2s^2 2p^6 3s^2 3p^3$1s^2 2s^2 2p^6 3s^2 3p^3$) has a half-filled, exceptionally stable 3p$3p$ subshell compared to Sulfur (3s^2 3p^4$3s^2 3p^4$). This makes it harder to remove an electron from P than from S. ### Step 2: Final Trend Construction Because of this half-filled stability, the ionization energy of P is greater than that of S. Therefore, the corrected trend becomes: Al < Si < S < P < Cl$Al < Si < S < P < Cl$ ### Pattern Recognition Always look for Group 15 (half-filled np^3$np^3$) vs Group 16 (np^4$np^4$) anomalies. Group 15 always has a higher first ionization energy than Group 16 in the same period. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Related Formula IE_2 text requires breaking stable noble gas configurations if M^+ text is isoelectronic with a noble gas.$$IE_2 \text{ requires breaking stable noble gas configurations if } M^+ \text{ is isoelectronic with a noble gas.}$$ ### Core Logic Statement I: Let's analyze the electronic configurations. Na$Na$ (Z=11$Z=11$): 1s^2 2s^2 2p^6 3s^1 implies Na^+$1s^2 2s^2 2p^6 3s^1 \implies Na^+$ is 1s^2 2s^2 2p^6$1s^2 2s^2 2p^6$ (Stable Neon noble gas core). Mg$Mg$ (Z=12$Z=12$): 1s^2 2s^2 2p^6 3s^2 implies Mg^+$1s^2 2s^2 2p^6 3s^2 \implies Mg^+$ is 1s^2 2s^2 2p^6 3s^1$1s^2 2s^2 2p^6 3s^1$. Removing a second electron from Na^+$Na^+$ (IE_2$IE_2$) involves disrupting a highly stable, fully-filled 2p^6$2p^6$ shell, requiring massive energy. Removing a second electron from Mg^+$Mg^+$ (IE_2$IE_2$) just removes the 3s^1$3s^1$ electron. Thus, IE_2$IE_2$ of Na > IE_2$IE_2$ of Mg. Statement I is true. Statement II: Both O^2-$O^{2-}$ and F^-$F^{-}$ are isoelectronic species, possessing 10 electrons (1s^2 2s^2 2p^6$1s^2 2s^2 2p^6$). However, the nuclear charge (number of protons, Z) is different. O^2-$O^{2-}$ has 8 protons pulling 10 electrons. F^-$F^{-}$ has 9 protons pulling 10 electrons. Since F^-$F^{-}$ has a higher effective nuclear charge (Z_texteff$Z_{\text{eff}}$), its electron cloud is pulled more tightly, making its radius smaller. Thus, the radius of O^2-$O^{2-}$ > F^-$F^{-}$. Statement II is true. ### Pattern Recognition For isoelectronic species, more negative charge always equals a larger ionic radius (lower Z/e$Z/e$ ratio implies less nuclear pull per electron). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity