MHT CET · Chemistry · p Block Elements (Group 15, 16, 17 & 18)

Which of the following is correct decreasing order of boiling point of compounds?

A \(\mathrm{CH}_3 \mathrm{Cl}\gt\mathrm{CH}_3 \mathrm{Br}\gt\mathrm{CH}_2 \mathrm{Br}_2\gt\mathrm{CHBr}_3\)
B \(\mathrm{CH}_3 \mathrm{Br}\gt\mathrm{CH}_2 \mathrm{Br}_2\gt\mathrm{CHBr}_3\gt\mathrm{CH}_3 \mathrm{Cl}\)
C \(\mathrm{CHBr}_3\gt\mathrm{CH}_2 \mathrm{Br}_2\gt\mathrm{CH}_3 \mathrm{Br}\gt\mathrm{CH}_3 \mathrm{Cl}\)
D \(\mathrm{CH}_3 \mathrm{Br}\gt\mathrm{CH}_3 \mathrm{Cl}\gt\mathrm{CHBr}_3\gt\mathrm{CH}_2 \mathrm{Br}_2\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{CHBr}_3\gt\mathrm{CH}_2 \mathrm{Br}_2\gt\mathrm{CH}_3 \mathrm{Br}\gt\mathrm{CH}_3 \mathrm{Cl}\)

Step-by-step Solution

Detailed explanation

The comparative boiling points of halogen derivatives are related to van der Waals forces of attraction which depend upon molecular size. The molecular size depends upon the size of halogen and number of halogen atoms. Thus, for the given compounds, boiling point decreases as, \(\mathrm{CHBr}_3\gt\mathrm{CH}_2 \mathrm{Br}_2\gt\mathrm{CH}_3 \mathrm{Br}\gt\mathrm{CH}_3 \mathrm{Cl}\).

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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