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Chapter 11: Problem 112

Select the substance in each pair that should have the higher boiling point. In each case identify the principal intermolecular forces involved and account briefly for your choice. (a) \(\mathrm{K}_{2} \mathrm{~S}\) or \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N},\) (b) \(\mathrm{Br}_{2}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Short Answer

Expert verified
For pair (a), \(\mathrm{K}_{2} \mathrm{~S}\) should have the higher boiling point primarily due to its stronger ionic bonds. For pair (b), \(\mathrm{Br}_{2}\) should have the higher boiling point due to its stronger London dispersion forces

Step by step solution

01

Compare the substances in pair (a)

Let's start by comparing \(\mathrm{K}_{2} \mathrm{S}\) and $\left(\mathrm{CH}_{3}\right)_{3}\mathrm{N}$. \(\mathrm{K}_{2} \mathrm{~S}\) is an ionic compound and thus its principal intermolecular force would be the ionic bond, which is typically stronger than any type of intermolecular force present in covalent compounds. Contrarily, $(\mathrm{CH}_{3})_{3} N$ is a covalent compound with dipole-dipole forces as its primary intermolecular force since the nitrogen atom makes the molecule polar. Thus, \(\mathrm{K}_{2} \mathrm{~S}\) should have the higher boiling point.
02

Compare the substances in pair (b)

Now let's compare \(\mathrm{Br}_{2}\) and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$. Both are covalent compounds and the intermolecular forces are London dispersion forces (induced dipole-dipole interaction) for both. However, \(\mathrm{Br}_{2}\) is a much larger molecule with more electrons, leading to stronger dispersion forces. Thus, \(\mathrm{Br}_{2}\) should have the higher boiling point.

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