Arrange the following haloalkanes in decreasing order of boiling point:

Chloromethane, Bromomethane, Dibromomethane, Bromoform

Bromoform > Dibromomethane > Bromomethane > Chloromethane

The correct answer is option 1. Bromoform > Dibromomethane > Bromomethane > Chloromethane.

The correct arrangement of the haloalkanes in decreasing order of boiling point is: Bromoform > Dibromomethane > Bromomethane > Chloromethane

Here's the reasoning:

Boiling point increases with increasing molecular weight: Generally, heavier molecules have stronger London dispersion forces (intermolecular forces) between them, requiring more energy to overcome and transition to the gas phase. Therefore, heavier molecules have higher boiling points.

Among the given compounds:

1. Chloromethane (CH₃Cl) has the lowest molecular weight (50.49 g/mol).
2. Bromomethane (CH₃Br) has a higher molecular weight (94.94 g/mol) due to the heavier bromine atom compared to chlorine.
3. Dibromomethane (CH₂Br₂) has an even higher molecular weight (173.82 g/mol) due to two bromine atoms.
4. Bromoform (CHBr₃) has the highest molecular weight (252.77 g/mol) with three bromine atoms.

Following this trend, bromoform will have the highest boiling point, followed by dibromethane, bromomethane, and then chloromethane.