The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715 kJ mol-1 respectively. The above…

Step 1: Identify elements A and B

Group 14 elements: C, Si, Ge, Sn, Pb

First ionisation enthalpies (kJ/mol): C(1086) > Si(786) > Ge(762) > Sn(708) > Pb(715)

The two lowest IE values in Group 14 are Sn (708) and Pb (715).

So: A = Sn (IE = 708), B = Pb (IE = 715)

Step 2: Analyse $\text{A}^{2+}$ = $\ce{Sn^{2+}}$

$\ce{Sn^{2+}}$ can be oxidised to $\ce{Sn^{4+}}$ (loses 2 more electrons) → acts as a reducing agent

$\ce{Sn^{2+} -> Sn^{4+} + 2e-}$ (reducing)

Step 3: Analyse $\text{B}^{4+}$ = $\ce{Pb^{4+}}$

$\ce{Pb^{4+}}$ tends to gain electrons to form more stable $\ce{Pb^{2+}}$ → acts as an oxidising agent

$\ce{Pb^{4+} + 2e- -> Pb^{2+}}$ (oxidising)

Answer: (c) reducing and oxidising