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  • Arrange the following in increasing order of their boiling pointOption: 1 <img alt="\begin{aligned} & \mathrm{lCl}>\mathrm{Cl}_2 \\ \end{align

Arrange the following in increasing order of their boiling point

Option: 1

\begin{aligned} & \mathrm{lCl}>\mathrm{Cl}_2 \\ \end{aligned}


Option: 2

\begin{aligned} & \mathrm{lCl}<\mathrm{Cl}_2 \\ \end{aligned}


Option: 3

\begin{aligned} & \mathrm{lCl}=\mathrm{Cl}_2 \end{aligned}


Option: 4

Can't be predicted


Answers (1)

best_answer

\mathrm{ICl} has dipole-dipole interaction whereas \mathrm{Cl}_2 has London forces which is weaker compared to the former. And boiling point is directly proportional to the strength of Vanderwall forces.

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