The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715mathrm\ kJ\ mol^-1$715\mathrm{\ kJ\ mol}^{-1}$ respectively. The above values are lowest among their group members. The nature of their ions mathrmA^2+$\mathrm{A}^{2+}$, mathrmB^4+$\mathrm{B}^{4+}$ respectively is:

### Core Logic For Group 14 (textC, textSi, textGe, textSn, textPb$\text{C}, \text{Si}, \text{Ge}, \text{Sn}, \text{Pb}$): - The ionisation energies generally decrease down the group, but there is an anomaly between textSn$\text{Sn}$ and textPb$\text{Pb}$ due to relativistic contraction / poor shielding of 4f electrons in textPb$\text{Pb}$. - Thus, the first ionisation enthalpy of Tin (mathrmSn$\mathrm{Sn}$) is 708 text kJ mol^-1$708 \text{ kJ mol}^{-1}$ and Lead (mathrmPb$\mathrm{Pb}$) is 715 text kJ mol^-1$715 \text{ kJ mol}^{-1}$. These are indeed the lowest in the group. - Hence, element **A** is mathrmSn$\mathrm{Sn}$ and **B** is mathrmPb$\mathrm{Pb}$. Nature of their ions: - mathrmA^2+ = mathrmSn^2+$\mathrm{A}^{2+} = \mathrm{Sn}^{2+}$: Since mathrmSn^4+$\mathrm{Sn}^{4+}$ is more stable than mathrmSn^2+$\mathrm{Sn}^{2+}$, mathrmSn^2+$\mathrm{Sn}^{2+}$ readily undergoes oxidation to +4$+4$, acting as a strong **reducing agent**. - mathrmB^4+ = mathrmPb^4+$\mathrm{B}^{4+} = \mathrm{Pb}^{4+}$: Due to the strong **inert pair effect**, mathrmPb^2+$\mathrm{Pb}^{2+}$ is highly stable compared to mathrmPb^4+$\mathrm{Pb}^{4+}$. Thus, mathrmPb^4+$\mathrm{Pb}^{4+}$ is eager to reduce to +2$+2$, acting as a strong **oxidising agent**. ### Pattern Recognition Inert pair effect becomes extremely prominent at the bottom of the group. Lead's most stable state is +2$+2$, making mathrmPb^4+$\mathrm{Pb}^{4+}$ oxidising. Tin's stable state is +4$+4$, making mathrmSn^2+$\mathrm{Sn}^{2+}$ reducing. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: p-Block Elements Class 11 Chemistry: Periodic Classification of Elements