To solve this problem, we need to understand the concept of the inert pair effect.$\newline$The inert pair effect refers to the reluctance of the $ns^2$ electrons in the valence shell of heavier elements$\newline$to participate in bonding. This effect is more pronounced in heavier elements of a group.$\newline$In the case of Group 14 elements, such as Sn and Pb:$\newline$• Sn (tin) and Pb (lead) can exhibit oxidation states of $+2$ and $+4. \newline$• The $+2$ oxidation state becomes more stable as we move down the group due to the inert pair effect.$\newline$• For Sn, the $+4$ oxidation state is more stable than the $+2$ state, making $Sn^{2+}$ a reducing agent.$\newline$• For Pb, the $+2$ oxidation state is more stable than the $+4$ state, making $Pb^{4+}$ an oxidizing agent.$\newline$Analyzing the options:$\newline$• Option 1: Incorrect, as $Sn^{2+}$ is reducing, not oxidizing.$\newline$• Option 2: Incorrect, as both ions cannot be both oxidizing and reducing simultaneously.$\newline$• Option 3: Incorrect, as $Sn^{4+}$ is not reducing.$\newline$• Option 4: Correct, as $Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidizing.$\newline$Therefore, the correct option is Option 4.