### Related Formula textIonic Radius propto frac1textNuclear Charge (Z) quad text(for Isoelectronic series)$\text{Ionic Radius } \propto \frac{1}{\text{Nuclear Charge } (Z)} \quad \text{(for Isoelectronic series)}$ ### Core Logic Evaluating each statement systematically : * Statement (I) is correct: mathrmMg^2+, mathrmNa^+, mathrmF^-, mathrmO^2-$\mathrm{Mg}^{2+}, \mathrm{Na}^{+}, \mathrm{F}^{-}, \mathrm{O}^{2-}$ all possess exactly 10 electrons (isoelectronic). As the positive nuclear charge decreases (Z = 12$Z = 12$ for mathrmMg$\mathrm{Mg}$ down to Z = 8$Z = 8$ for mathrmO$\mathrm{O}$), the nucleus exerts less pull on the electron cloud, causing the ionic radius to increase : mathrmMg^2+ < mathrmNa^+ < mathrmF^- < mathrmO^2-$\mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-} < \mathrm{O}^{2-}$ * Statement (II) is correct: Chlorine has a higher electron gain enthalpy magnitude than fluorine due to lower electron-electron repulsion in its larger 3p$3p$ orbital. The standard halogen trend follows: mathrmCl > mathrmF > mathrmBr > mathrmI$\mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I}$ Thus, both statements are correct. ### Pattern Recognition For species with the same number of electrons, a higher negative charge always leads to a larger electron cloud radius due to reduced nuclear traction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic F^-$F^-$, Ne$Ne$, Na^+$Na^+$ all have 1s^2, 2s^2, 2p^6$1s^2, 2s^2, 2p^6$ configuration (10 electrons). However, their atomic numbers (nuclear charge, Z$Z$) are different: F$F$: Z = 9$Z = 9$ Ne$Ne$: Z = 10$Z = 10$ Na$Na$: Z = 11$Z = 11$ Because they have the same number of electrons but different nuclear charges, the attraction between the nucleus and the valence shell electrons will differ. ### Step 1: Final Conclusion Higher nuclear charge strongly attracts the isoelectronic electron cloud, decreasing the ionic radius. Hence, their size is primarily affected by the nuclear charge (z$z$). ### Pattern Recognition For isoelectronic species, size is inversely proportional to atomic number Z$Z$. The greater the Z$Z$, the smaller the size. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### Core Logic Let's classify the nature of each given oxide: - Cl_2O_7$Cl_2O_7$: Non-metal oxide in highest oxidation state rightarrow$\rightarrow$ Strongly Acidic. - CO$CO$: Neutral oxide. - PbO_2$PbO_2$: Heavy metal oxide near metalloid line rightarrow$\rightarrow$ Amphoteric. - N_2O$N_2O$: Neutral oxide. - NO$NO$: Neutral oxide. - Al_2O_3$Al_2O_3$: Classic amphoteric oxide. - SiO_2$SiO_2$: Weakly acidic oxide. - N_2O_5$N_2O_5$: Non-metal oxide rightarrow$\rightarrow$ Acidic. - SnO_2$SnO_2$: Heavy metal oxide near metalloid line rightarrow$\rightarrow$ Amphoteric. ### Step 1: Count Amphoteric Oxides The amphoteric oxides in the list are Al_2O_3$Al_2O_3$, SnO_2$SnO_2$, and PbO_2$PbO_2$. Total count = 3. ### Pattern Recognition Memorize the main neutral oxides (N_2O, NO, CO$N_2O, NO, CO$) and the classic amphoteric oxides (Al_2O_3, ZnO, PbO, PbO_2, SnO, SnO_2, BeO, As_2O_3, Sb_2O_3$Al_2O_3, ZnO, PbO, PbO_2, SnO, SnO_2, BeO, As_2O_3, Sb_2O_3$). High oxidation state non-metals are always acidic. ### 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 textIonization Enthalpy (IE_1) propto frac1, textAtomic Size quad textand quad textStable configuration enhancements.$\text{Ionization Enthalpy } (IE_1) \propto \frac{1}, {\text{Atomic Size}} \quad \text{and} \quad \text{Stable configuration enhancements.}$ ### Core Logic Analyzing periodic trends: 1. Ionization energy increases across a period from left to right and decreases down a group. 2. Nitrogen (N$N$) and Carbon (C$C$) belong to Period 2, while Aluminum (Al$Al$) and Silicon (Si$Si$) belong to Period 3. Consequently, Period 2 elements have smaller atomic radii and higher ionization energies. 3. Comparing Nitrogen and Carbon, Nitrogen (1s^2 2s^2 2p^3$1s^2 2s^2 2p^3$) has a highly stable, half-filled p$p$-subshell configuration, giving it a much higher ionization energy than Carbon. ### Step 1: Trend Layout The overall first ionization enthalpy trend follows the sequence: mathrmAl < mathrmSi < mathrmC < mathrmN$\mathrm{Al} < \mathrm{Si} < \mathrm{C} < \mathrm{N}$ ### Pattern Recognition Nitrogen exhibits an exceptionally high first ionization energy due to its small size combined with a stable, half-filled 2p^3$2p^3$ valence subshell. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity

### Related Formula Electron Gain Enthalpy (Delta_egH$\Delta_{eg}H$) trends in Group 16 and Group 17 elements. ### Core Logic Evaluating both statements: * **Statement I is false**: Due to its small size, strong inter-electronic repulsions in the compact 2p$2p$ subshell of Fluorine limit incoming electrons compared to Chlorine. As a result, Chlorine (textCl$\text{Cl}$) has the *most negative* electron gain enthalpy in Group 17. * **Statement II is true**: Similarly, the exceptionally small size of the Oxygen atom creates intense electron-electron repulsions. Consequently, it has the *least negative* electron gain enthalpy among all elements in Group 16. ### Step 1: Final Assessment Therefore, Statement I is false while Statement II is true, matching choice (4). ### Pattern Recognition Third-period elements (textCl$\text{Cl}$, textS$\text{S}$) exhibit more negative electron gain enthalpies than their second-period counterparts (textF$\text{F}$, textO$\text{O}$) due to lower inter-electronic repulsion in their larger valence shells. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity