### Core Logic Let's analyze the first ionization energy (textIE_1$\text{IE}_1$) values for Group 14 elements (mathrmC, Si, Ge, Sn, Pb$\mathrm{C, Si, Ge, Sn, Pb}$): Generally, ionization energy decreases down a group as atomic size increases. However, heavy post-transition elements exhibit an anomaly: * Analysis of Statement I: Moving from mathrmSn$\mathrm{Sn}$ to mathrmPb$\mathrm{Pb}$, the 4f$4f$ orbital subshell becomes fully filled. Because 4f$4f$ electrons provide very poor shielding, the outer valence electrons experience a significantly higher effective nuclear charge (Z_texteff$Z_{\text{eff}}$). This inert pair effect contractive behavior tightly binds the outer electrons, making the first ionization energy of Lead higher than that of Tin: textIE_1(mathrmPb) = 715text kJ/mol > textIE_1(mathrmSn) = 708text kJ/mol$\text{IE}_1(\mathrm{Pb}) = 715\text{ kJ/mol} > \text{IE}_1(\mathrm{Sn}) = 708\text{ kJ/mol}$ Thus, Statement I is true. * Analysis of Statement II: Following normal periodic trends down the group, the ionization energy decreases from Silicon to Germanium due to the increasing atomic radius: textIE_1(mathrmSi) = 786text kJ/mol > textIE_1(mathrmGe) = 761text kJ/mol$\text{IE}_1(\mathrm{Si}) = 786\text{ kJ/mol} > \text{IE}_1(\mathrm{Ge}) = 761\text{ kJ/mol}$ Therefore, the claim that mathrmGe > mathrmSi$\mathrm{Ge} > \mathrm{Si}$ is false. ### Pattern Recognition The overall first ionization energy trend for Group 14 is: mathrmC > Si > Ge > Pb > Sn$\mathrm{C > Si > Ge > Pb > Sn}$. Notice that Lead breaks the downward trend and has a higher ionization energy than Tin due to poor shielding by 4f$4f$ electrons (lanthanide contraction). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Classification of Elements and Periodicity in Properties Class 11 Chemistry: The p-Block Elements