To solve this problem, we need to determine the molecular geometry of each compound listed in List-I and match it with the correct geometry in List-II.$\\$Let's analyze each compound:$\\$(a) $PCl_5: \\$• $PCl_5$ has a central phosphorus atom surrounded by five chlorine atoms.$\\$• The electron geometry is based on five bonding pairs, which is $$trigonal bipyramidal$. \\$• Therefore, the molecular geometry of $PCl_5$ is $$trigonal bipyramidal$. \\$(b) $SF_6: \\$• $SF_6$ has a central sulfur atom surrounded by six fluorine atoms.$\\$• The electron geometry is based on six bonding pairs, which is $$octahedral$. \\$• Therefore, the molecular geometry of $SF_6$ is $$octahedral$. \\$(c) $BrF_5: \\$• $BrF_5$ has a central bromine atom surrounded by five fluorine atoms and one lone pair.$\\$• The electron geometry is based on six regions of electron density, which is $$octahedral$. \\$• With one lone pair, the molecular geometry becomes $$square pyramidal$. \\$(d) $BF_3: \\$• $BF_3$ has a central boron atom surrounded by three fluorine atoms.$\\$• The electron geometry is based on three bonding pairs, which is $$trigonal planar$. \\$• Therefore, the molecular geometry of $BF_3$ is $$trigonal planar$. \\$Now, let's match each compound with its corresponding geometry:$\\$• $(a) PCl_5 \rightarrow$(iv) Trigonal bipyramidal$\\$• $(b) SF_6 \rightarrow$(iii) Octahedral$\\$• $(c) BrF_5 \rightarrow$(i) Square pyramidal$\\$• $(d) BF_3 \rightarrow$(ii) Trigonal planar$\\$Therefore, the correct option is Option 1:$\\$(a) $\rightarrow$(iv), (b) $\rightarrow$(iii), (c) $\rightarrow$(i), (d) $\rightarrow$(ii)$\\$