Methyl iodide (CH3I) has a higher boiling point than methyl fluoride (CH3F) due to the differences in their intermolecular forces.
Methyl iodide is a larger molecule than methyl fluoride, which means it has more electrons and therefore a greater London dispersion force (also known as van der Waals force). This force results from the temporary dipoles that form in molecules due to the fluctuation of electrons around their nuclei. As a result, methyl iodide molecules are held together more tightly, requiring more energy to overcome the intermolecular forces and boil.
On the other hand, methyl fluoride has a smaller molecular size and a weaker London dispersion force compared to methyl iodide, resulting in a lower boiling point. Additionally, methyl fluoride has a dipole moment due to the electronegativity difference between carbon and fluorine atoms, leading to a weak dipole-dipole interaction. However, this dipole-dipole interaction is not as strong as the London dispersion forces in methyl iodide.
Therefore, the stronger intermolecular forces in methyl iodide result in a higher boiling point than in methyl fluoride.