Electronic configuration of four elements A, B, C and D are given below : (A) 1mathrms^2 2mathrms^2 2mathrmp^3$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^3$ (B) 1mathrms^2 2mathrms^2 2mathrmp^4$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^4$ (C) 1mathrms^2 2mathrms^2 2mathrmp^5$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^5$ (D) 1mathrms^2 2mathrms^2 2mathrmp^2$1\mathrm{s}^2 2\mathrm{s}^2 2\mathrm{p}^2$ Which of the following is the correct order of increasing electronegativity (Pauling's scale)?

### Related Formula textIonic Character propto lvert Delta H_textIE - Delta H_textEGE rvert$$\text{Ionic Character} \propto \lvert \Delta H_{\text{IE}} - \Delta H_{\text{EGE}} \rvert$$ ### Core Logic The relative ionic quality of a standard binary bond rises as the gap scale between ionization enthalpy and negative electron gain enthalpy parameters widens. Comparing the values: * For B: Delta H_textEGE = -349 mathrm~kJ/mol$\Delta H_{\text{EGE}} = -349 \mathrm{~kJ/mol}$ (largest energy release) rightarrow$\rightarrow$ Highest ionic character. * For A: Delta H_textEGE = -328 mathrm~kJ/mol$\Delta H_{\text{EGE}} = -328 \mathrm{~kJ/mol}$. * For C: Delta H_textEGE = -325 mathrm~kJ/mol$\Delta H_{\text{EGE}} = -325 \mathrm{~kJ/mol}$. * For D: Delta H_textEGE = -295 mathrm~kJ/mol$\Delta H_{\text{EGE}} = -295 \mathrm{~kJ/mol}$ (smallest energy release) rightarrow$\rightarrow$ Lowest ionic character. Arranging them in descending order of ionic character yields: mathrmEB gt mathrmEA gt mathrmEC gt mathrmED$$\mathrm{EB} \gt \mathrm{EA} \gt \mathrm{EC} \gt \mathrm{ED}$$ ### Pattern Recognition A highly exothermic electron gain enthalpy value favors easier anion production, widening electronegativity variations to enhance ionic bond properties. ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties

### 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