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. K₂O > Na₂O > MgO > Al₂O₃$K_{2}O > Na_{2}O > MgO > Al_{2}O_{3}$ – Basic character Choose the correct answer from the option given below.

Core Logic

• Statement I: Correct order of size is Mg > Al > Mg²⁺ > Al³⁺$\text{Mg} > \text{Al} > \text{Mg}^{2+} > \text{Al}^{3+}$ 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 Cl > Br > S > O$\text{Cl} > \text{Br} > \text{S} > \text{O}$ 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

Core Logic

Step 1: Evaluate Statement-I:

• Across Period 2, IE₁$IE_1$ generally increases with Zeff$Z_{\text{eff}}$.
• N (2p³$2p^3$) is half-filled, so IE₁(N) > IE₁(O)$IE_1(\text{N}) > IE_1(\text{O})$.
• Correct order: C < O < N < F$\text{C} < \text{O} < \text{N} < \text{F}$. Statement-I is TRUE.

Step 2: Evaluate Statement-II:

• Magnitudes of ΔegH$\Delta_{eg}H$ for Group 16: S (200) > Se (195) > Te (190) > Po (174) > O (141 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 IE₁$IE_1$, and small 2p shell makes O < Po for |ΔegH|$|\Delta_{eg}H|$.

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 (Zeff$Z_{\text{eff}}$). 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 (1s² 2s² 2p⁶ 3s² 3p³$1s^2 2s^2 2p^6 3s^2 3p^3$) has a half-filled, exceptionally stable 3p$3p$ subshell compared to Sulfur (3s² 3p⁴$3s^2 3p^4$). 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 np³$np^3$) vs Group 16 (np⁴$np^4$) anomalies. Group 15 always has a higher first ionization energy than Group 16 in the same period.

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity in Properties

Related Formula

Core Logic

Statement I: Let's analyze the electronic configurations. Na$Na$ (Z=11$Z=11$): 1s² 2s² 2p⁶ 3s¹ Na^+$1s^2 2s^2 2p^6 3s^1 \implies Na^+$ is 1s² 2s² 2p⁶$1s^2 2s^2 2p^6$ (Stable Neon noble gas core). Mg$Mg$ (Z=12$Z=12$): 1s² 2s² 2p⁶ 3s² Mg^+$1s^2 2s^2 2p^6 3s^2 \implies Mg^+$ is 1s² 2s² 2p⁶ 3s¹$1s^2 2s^2 2p^6 3s^1$. Removing a second electron from Na^+$Na^+$ (IE₂$IE_2$) involves disrupting a highly stable, fully-filled 2p⁶$2p^6$ shell, requiring massive energy. Removing a second electron from Mg^+$Mg^+$ (IE₂$IE_2$) just removes the 3s¹$3s^1$ electron. Thus, IE₂$IE_2$ of Na > IE₂$IE_2$ of Mg. Statement I is true.

Statement II: Both O²⁻$O^{2-}$ and F⁻$F^{-}$ are isoelectronic species, possessing 10 electrons (1s² 2s² 2p⁶$1s^2 2s^2 2p^6$). However, the nuclear charge (number of protons, Z) is different. O²⁻$O^{2-}$ has 8 protons pulling 10 electrons. F⁻$F^{-}$ has 9 protons pulling 10 electrons. Since F⁻$F^{-}$ has a higher effective nuclear charge (Zeff$Z_{\text{eff}}$), its electron cloud is pulled more tightly, making its radius smaller. Thus, the radius of O²⁻$O^{2-}$ > 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).

Chapter Mix

Class 11 Chemistry: Classification of Elements and Periodicity