Q13E On the basis of intermolecular a... [FREE SOLUTION]

Chapter 10: Q13E (page 584)

On the basis of intermolecular attractions, explain the differences in the boiling points of n-butane \(\left( { - 1 ^\circ C} \right)\) and chloroethane \(\left( {12 ^\circ C} \right),\)which have similar molar masses.

Short Answer

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Chloroethane has a higher boiling point than \({\rm{n}} - butane\) although they have similar masses.

Step by step solution

Types of Intermolecular Forces

Hydrogen bondingis an example of a dipole-dipole force. This is an electrostatic interaction between the permanent dipoles in molecules.

Ion-dipole forcesare electrostatic interactions between one molecule's partially charged dipole and a fully charged ion.

The weakest of the intermolecular interactions鈥攊nstantaneous dipole-induced dipole forces鈥攁lso known as the London dispersion forces, are categorized as van der Waals forces. The correlated electron movements in the interacting molecules produce the Vander walls forces .

Explanation for the difference in the strength of dipoles

The difference in the strengths of the dipoles leads to one of the compounds having a higher boiling point than the other compound.

The \({\rm{C}} - {\rm{H}}\)bond is less polar than the \({\rm{C}} - {\rm{Cl}}\) bond, and as a result, the extent of the dipole interaction is more in the chloroethane as compared to the \({\rm{n}} - butane\). Therefore, although both chloroethane and \({\rm{n}} - butane\) have similar masses, the former has a higher boiling point than the latter.