Step 1: Understanding the Question:
We are given four different methyl halides and asked to identify the one with the lowest boiling point.
Step 2: Key Formula or Approach:
For a given alkyl group (in this case, the methyl group), the boiling point of the alkyl halide depends on the magnitude of intermolecular van der Waals forces (specifically London dispersion forces and dipole-dipole interactions).
These forces increase as the size and molecular mass of the halogen atom increase.
Step 3: Detailed Explanation:
Let's look at the halogens involved: Fluorine (F), Chlorine (Cl), Bromine (Br), and Iodine (I).
As we go down Group 17 in the periodic table, atomic mass and atomic size increase in the order: $\text{F} < \text{Cl} < \text{Br} < \text{I}$.
Therefore, the magnitude of van der Waals forces and the corresponding boiling points of the methyl halides follow the same trend:
$$\text{CH}_3\text{F} < \text{CH}_3\text{Cl} < \text{CH}_3\text{Br} < \text{CH}_3\text{I}$$
Fluoromethane ($\text{CH}_3\text{F}$) has the smallest halogen atom, the lowest molecular mass, and therefore the weakest intermolecular forces among the choices.
Step 4: Final Answer:
Fluoromethane has the lowest boiling point, matching option (A).