To determine which molecule has the maximum 'lone pair-lone pair' electron repulsions, we need to analyze the molecular geometry and the number of lone pairs present in each molecule.$\\$Let's examine each option:$\\$• $$ClF$_3: \\$Chlorine trifluoride has a $$T-shaped$$ molecular geometry.$\\$It has 5 regions of electron density: 3 bonding pairs and 2 lone pairs.$\\$• $$IF$_5: \\$Iodine pentafluoride has a $$square pyramidal$$ molecular geometry.$\\$It has 6 regions of electron density: 5 bonding pairs and 1 lone pair.$\\$• $$SF$_4: \\$Sulfur tetrafluoride has a $$see-saw$$ molecular geometry.$\\$It has 5 regions of electron density: 4 bonding pairs and 1 lone pair.$\\$• $$XeF$_2: \\$Xenon difluoride has a $$linear$$ molecular geometry.$\\$It has 5 regions of electron density: 2 bonding pairs and 3 lone pairs.$\\$Analysis:$\\$The number of lone pairs is crucial for determining the extent of lone pair-lone pair repulsions.$\\$Among the given molecules, $$XeF$_2$ has the highest number of lone pairs, which is 3.$\\$These lone pairs are arranged in a trigonal bipyramidal electron geometry,$\\$with the lone pairs occupying the equatorial positions to minimize repulsion.$\\$Conclusion:$\\$The molecule with the maximum 'lone pair-lone pair' electron repulsions is $$XeF$_2. \\$Therefore, the correct option is 4.