### 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