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

<p>Given the ionization energy values for elements $A$ and $B$ in group 14, where $A$ has an ionization energy of 708 kJ/mol and $B$ has 715 kJ/mol, these values are the lowest among their group. This helps identify the elements as tin (Sn) and lead (Pb), respectively. In this context:</p> <p><p>The $\mathrm{A}^{2+}$ ion corresponds to $\mathrm{Sn}^{2+}$, which acts as a reducing agent.</p></p> <p><p>The $\mathrm{B}^{4+}$ ion corresponds to $\mathrm{Pb}^{4+}$, which acts as an oxidizing agent.</p></p> <p>Therefore, Sn in the +2 oxidation state tends to donate electrons (reduce), while Pb in the +4 oxidation state tends to accept electrons (oxidize), underscoring their respective roles as reducing and oxidizing agents.</p>