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

The given first-ionisation enthalpies are 708 and 715 $$\text{kJ mol}^{-1}$$ for the two Group 14 elements $$A$$ and $$B$$ respectively.

Within Group 14 the experimental values are approximately:
$$C : 1086,\; Si : 786,\; Ge : 762,\; Sn : 708,\; Pb : 715\; \text{kJ mol}^{-1}$$

Hence
$$A = Sn\;(708\; \text{kJ mol}^{-1}), \qquad B = Pb\;(715\; \text{kJ mol}^{-1})$$

The ions mentioned are $$A^{2+} = Sn^{2+}$$ and $$B^{4+} = Pb^{4+}$$.

Sn$$^{2+}$$
Sn belongs to the 5$$s^{2}$$5$$p^{2}$$ family. The +2 state loses only the two 5$$p$$ electrons, leaving the inert pair 5$$s^{2}$$ unused. Sn$$^{2+}$$ is less stable than Sn$$^{4+}$$. Therefore Sn$$^{2+}$$ readily gets oxidised to Sn$$^{4+}$$ and, in the process, reduces other species. Hence Sn$$^{2+}$$ behaves as a reducing agent.

Pb$$^{4+}$$
Down the group the inert-pair effect becomes stronger. For lead the 6$$s^{2}$$ pair is very reluctant to participate in bonding, so the +2 state (Pb$$^{2+}$$) is far more stable than the +4 state. Pb$$^{4+}$$ therefore tends to accept electrons and get reduced to Pb$$^{2+}$$, i.e. it oxidises other species. Hence Pb$$^{4+}$$ acts as an oxidising agent.

Therefore, the nature of the two ions is:
$$Sn^{2+}$$ - reducing, $$Pb^{4+}$$ - oxidising.

The correct choice is Option C: reducing and oxidising.