Q.Arrange $\text{CH}_3\text{CH}_2\text{F}$, $\text{CH}_3\text{CH}_2\text{Cl}$, $\text{CH}_3\text{CH}_2\text{Br}$ and $\text{CH}_3\text{CH}_2\text{I}$ in increasing order of boiling point, and explain the trend.

Haloalkanes boil higher than the parent alkane of the same skeleton, because the polar $\text{C--X}$ bond adds dipole-dipole attraction on top of existing dispersion forces, and because the halogen atom, especially a heavier one, is markedly more polarisable than the hydrogen it replaces. Among the methyl halides this drives a boiling-point order $\text{CH}_3\text{F} < \text{CH}_3\text{Cl} < \text{CH}_3\text{Br} < \text{CH}_3\text{I}$ that rises with halogen size/polarisability even though bond polarity itself falls in the same direction -- polarisability, not polarity, dominates the trend. As with plain alkanes, increased branching lowers boiling point for a given molecular formula, since a more compact, spherical shape reduces intermolecular surface contact. Haloalkanes and haloarenes dissolve poorly in water (which would require breaking water's hydrogen-bond network without an adequately strong replacement interaction) but dissolve readily in common organic solvents, and density increases with the number and mass of halogens present, so polyhalomethanes such as $\text{CHCl}_3$ and $\text{CCl}_4$ are denser than water.