Which one of the following molecules is polar?

To determine which molecule is polar, we need to recall that a polar molecule has a net dipole moment. This occurs when the molecule has polar bonds (due to electronegativity differences) and an asymmetric geometry that prevents the bond dipoles from canceling out. Let's analyze each option step by step.

Starting with Option A: $$XeF_4$$. Xenon (Xe) is the central atom with 8 valence electrons. It forms four bonds with fluorine atoms, using all 8 electrons in bonding. The molecule has six electron domains: four bonding pairs and two lone pairs. The electron domain geometry is octahedral, but the two lone pairs are positioned opposite each other, resulting in a square planar molecular geometry. In this symmetric arrangement, the bond dipoles cancel each other, leading to a net dipole moment of zero. Therefore, $$XeF_4$$ is non-polar.

Next, Option B: $$IF_5$$. Iodine (I) is the central atom with 7 valence electrons. Five fluorine atoms bond to iodine, forming five I-F bonds. The total valence electrons are calculated as: Iodine contributes 7 electrons, and each fluorine contributes 7, so total = 7 + 5 × 7 = 42 electrons. The five bonds account for 10 electrons (2 per bond), leaving 42 - 10 = 32 electrons. These are distributed as lone pairs: each fluorine has three lone pairs (6 electrons), so 5 fluorines × 6 electrons = 30 electrons. The remaining 2 electrons form a lone pair on iodine. Thus, iodine has five bonding pairs and one lone pair, giving six electron domains. The electron domain geometry is octahedral, but with one lone pair, the molecular geometry is square pyramidal. This asymmetric shape prevents the bond dipoles from canceling, resulting in a net dipole moment. Hence, $$IF_5$$ is polar.

Moving to Option C: $$SbF_5$$. Antimony (Sb) is the central atom with 5 valence electrons. Five fluorine atoms bond to antimony. Total valence electrons: Sb contributes 5, each F contributes 7, so total = 5 + 5 × 7 = 40 electrons. The five bonds use 10 electrons, leaving 40 - 10 = 30 electrons. These are assigned as lone pairs on fluorine atoms: each fluorine has three lone pairs (6 electrons), so 5 fluorines × 6 = 30 electrons. Antimony has no lone pairs. With five bonding pairs and no lone pairs, the electron domain geometry is trigonal bipyramidal. This symmetric arrangement cancels the bond dipoles, leading to a net dipole moment of zero. Therefore, $$SbF_5$$ is non-polar.

Finally, Option D: $$CF_4$$. Carbon (C) is the central atom with 4 valence electrons. Four fluorine atoms bond to carbon. Total valence electrons: C contributes 4, each F contributes 7, so total = 4 + 4 × 7 = 32 electrons. The four bonds use 8 electrons, leaving 32 - 8 = 24 electrons. Each fluorine has three lone pairs (6 electrons), so 4 fluorines × 6 = 24 electrons. Carbon has no lone pairs. With four bonding pairs, the electron domain geometry is tetrahedral. The symmetric arrangement of identical bonds cancels the bond dipoles, resulting in a net dipole moment of zero. Thus, $$CF_4$$ is non-polar.

In summary, only $$IF_5$$ (Option B) has an asymmetric geometry (square pyramidal) that makes it polar. Hence, the correct answer is Option B.