To solve this problem, we need to determine the number of Faradays required for each conversion in List I and match it with the correct option in List II.$$Let's analyze each conversion:$\\$A. Conversion of 1 mol of $H_2O$ to $O_2: \\$The balanced reaction is:$\\ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \\$This means 2 moles of $H_2O$ require 4 moles of electrons.$\\$Therefore, 1 mole of $H_2O$ requires 2 moles of electrons, or 2F.$\\$B. Conversion of 1 mol of $MnO_4^-$ to $Mn^{2+}: \\$The balanced reaction is:$\\ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \\$This means 1 mole of $MnO_4^-$ requires 5 moles of electrons, or 5F.$\\$C. Conversion of 1.5 mol of $Ca$ from molten $CaCl_2: \\$The balanced reaction is:$\\ Ca^{2+} + 2e^- \rightarrow Ca \\$This means 1 mole of $Ca$ requires 2 moles of electrons.$\\$Therefore, 1.5 moles of $Ca$ require 3 moles of electrons, or 3F.$\\$D. Conversion of 1 mol of $Fe_2O_3$ to $Fe: \\$The balanced reaction is:$\\ Fe_2O_3 + 6e^- \rightarrow 2Fe + 3O^{2-} \\$This means 1 mole of $Fe_2O_3$ requires 6 moles of electrons, or 6F.$\\$However, since we are considering the conversion to 1 mole of $Fe, \\$it requires 3 moles of electrons, or 3F.$\\$Now, let's match these with List II:$\\$A. 1 mol of $H_2O$ to $O_2 \rightarrow 2F \\$B. 1 mol of $MnO_4^-$ to $Mn^{2+} \rightarrow 5F \\$C. 1.5 mol of $Ca$ from molten $CaCl_2 \rightarrow 3F \\$D. 1 mol of $Fe_2O_3$ to $Fe \rightarrow 3F \\$Therefore, the correct matching is:$\\ A \rightarrow II, B \rightarrow IV, C \rightarrow I, D \rightarrow III \\$This corresponds to Option 3.