The polarity of a molecule is determined by its net dipole moment. A molecule is polar if it has polar bonds and its molecular geometry allows for a net dipole moment. In the given compounds (CH\u2082X\u2082, where X is a halogen), the central carbon atom is sp\u00b3 hybridized, leading to a tetrahedral geometry. Due to the presence of two hydrogen atoms and two halogen atoms, the molecule is unsymmetrical, and the bond dipoles do not cancel out, resulting in a net dipole moment.

The polarity of the C-X bond depends on the electronegativity difference between carbon and the halogen. The greater the electronegativity difference, the more polar the bond. The electronegativity order for halogens is F > Cl > Br > I. Therefore, the C-F bond is the most polar among C-F, C-Cl, C-Br, and C-I bonds.

While bond length also plays a role (dipole moment = charge \u00d7 distance), the electronegativity difference is the dominant factor in determining bond polarity for these molecules. Fluorine is the most electronegative element, leading to the largest electronegativity difference with carbon and thus the most polar C-F bond. This results in CH\u2082F\u2082 having the largest net dipole moment and being the most polar compound.

• A) CH\u2082I\u2082: Iodine has the lowest electronegativity among the halogens listed, leading to the least polar C-I bonds and thus the least polar compound.
• B) CH\u2082F\u2082: Fluorine is the most electronegative halogen, resulting in the most polar C-F bonds and the highest net dipole moment for the molecule.
• C) CH\u2082Cl\u2082: Chlorine is less electronegative than fluorine, so the C-Cl bonds are less polar than C-F bonds.
• D) CH\u2082Br\u2082: Bromine is less electronegative than both fluorine and chlorine, making the C-Br bonds less polar than C-F and C-Cl bonds.

Correct Answer: (B)