Dans chacune des séries suivantes, attribuez à chacun des...

Dans chacune des séries suivantes, attribuez à chacun des composés l'une des températures de vaporisation indiquées en ${ }^{\circ} \mathrm{C}$ : a) $\mathrm{HBr}, \mathrm{HCl}, \mathrm{HI}, \mathrm{HF} \quad-\quad 19^{\circ} \mathrm{C},-67^{\circ} \mathrm{C},-84^{\circ} \mathrm{C},-35^{\circ} \mathrm{C}$ b) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}, \mathrm{H}_2 \mathrm{C}=\mathrm{O}, \mathrm{CH}_3-\mathrm{OH}--21^{\circ} \mathrm{C}, 78{ }^{\circ} \mathrm{C}, 65^{\circ} \mathrm{C}$ c) $\mathrm{HO}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}, \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}-197^{\circ} \mathrm{C}, 97^{\circ} \mathrm{C}$ d) $\mathrm{CH}_3-\mathrm{OH}, \mathrm{CH}_3-\mathrm{SH}-7^{\circ} \mathrm{C}, 65^{\circ} \mathrm{C}$ e) $\mathrm{Xe}, \mathrm{Ne}, \mathrm{Ar}, \mathrm{Kr} \quad-\quad-153^{\circ} \mathrm{C},-246{ }^{\circ} \mathrm{C},-107^{\circ} \mathrm{C},-186^{\circ} \mathrm{C}$

Dans chacune des séries suivantes, attribuez à chacun des composés l'une des températures de vaporisation indiquées en ${ }^{\circ} \mathrm{C}$ :
a) $\mathrm{HBr}, \mathrm{HCl}, \mathrm{HI}, \mathrm{HF} \quad-\quad 19^{\circ} \mathrm{C},-67^{\circ} \mathrm{C},-84^{\circ} \mathrm{C},-35^{\circ} \mathrm{C}$
b) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OH}, \mathrm{H}_2 \mathrm{C}=\mathrm{O}, \mathrm{CH}_3-\mathrm{OH}--21^{\circ} \mathrm{C}, 78{ }^{\circ} \mathrm{C}, 65^{\circ} \mathrm{C}$
c) $\mathrm{HO}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}, \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}-197^{\circ} \mathrm{C}, 97^{\circ} \mathrm{C}$
d) $\mathrm{CH}_3-\mathrm{OH}, \mathrm{CH}_3-\mathrm{SH}-7^{\circ} \mathrm{C}, 65^{\circ} \mathrm{C}$
e) $\mathrm{Xe}, \mathrm{Ne}, \mathrm{Ar}, \mathrm{Kr} \quad-\quad-153^{\circ} \mathrm{C},-246{ }^{\circ} \mathrm{C},-107^{\circ} \mathrm{C},-186^{\circ} \mathrm{C}$

Key Concepts

Influence of Molecular Structure on Boiling Point

The boiling point of a compound is affected by its molecular structure, including its size, shape, and polarity. Larger molecules with greater surface area generally exhibit higher boiling points due to increased London dispersion forces, while the presence of polar groups and the capacity to form hydrogen bonds also significantly elevate boiling points by enhancing attractive intermolecular forces.

Dipole-Dipole Interactions

Dipole-dipole interactions occur between molecules that have permanent dipoles arising from differences in electronegativity between atoms. These interactions contribute to higher boiling points compared to molecules exhibiting only London dispersion forces, because the electrostatic attractions need more energy to disrupt.

London Dispersion Forces

London dispersion forces are weak intermolecular attractions arising from temporary fluctuations in electron density within molecules. Although these forces are generally weaker than hydrogen bonds and dipole-dipole interactions, they become significant in larger or more polarizable molecules and play an important role in influencing boiling points, particularly in nonpolar substances.

Intermolecular Forces

Intermolecular forces are the attractions between molecules, which include hydrogen bonding, dipole-dipole interactions, and London dispersion forces. These forces largely determine physical properties such as boiling points. Stronger intermolecular forces typically lead to higher boiling points because more energy is required to separate the molecules during vaporization.

Hydrogen Bonding

Hydrogen bonding is a particular strong type of dipole-dipole interaction that occurs when hydrogen is bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine. This bond significantly increases the boiling point of a compound, as it enhances the overall intermolecular attractions and requires more energy to break during phase changes.

Transcript

00:01 In each part of this problem, we are given three different molecules, and we want to determine which one out of these three best meets the given criterion in each part of the problem.

00:12 So starting in part a, we are given three molecules, and we want to determine which one of these has the highest boiling point.

00:20 For each part of this problem, we need to compare the relative strengths of the intermolecular forces present in each one of the given molecules in order to help us determine this.

00:31 So for a boiling point, we know that stronger intermolecular forces will require a higher boiling point.

00:40 And this is because more energy is required and therefore a higher temperature in order to overcome stronger intermolecular forces.

00:50 So if we compare the three compounds given in part a, we see that they are all non -polar and they all differ by the atom before of the same atom that, are bonded to the carbon atom.

01:04 If we look at the periodic table, we see that bromine compared to chlorine and fluorine has a higher molar mass.

01:12 And so therefore, cbr4 is the largest molecule out of these three, which leads to the highest polarizability.

01:19 So even though all of these display london dispersion forces, we can conclude that cbr4 has the strongest intermolecular forces out of these three, and therefore the highest boiling point.

01:31 In part b, we want to know which one of the three molecules has the lowest freezing point.

01:36 When we freeze something, we are going from a liquid to a solid.

01:41 And as we continue to decrease the temperature, the movement of the molecules also slows down as well.

01:49 And when we slow down those particles, we know that they pack together more tightly and resemble a solid.

01:59 And so therefore substances with higher intermolecular forces will resemble a solid more so than those with lower intermolecular forces.

02:10 And so that means that we are looking for the substance with the smallest intermolecular force to correspond to the lowest freezing point.

02:20 We know that lif is an ionic compound, so that is the strongest type of intermolecular force, and hcl is a polar molecule that displays dipole interact with.

02:33 However, f2 is a diatomic and therefore non -polar, and it only displays london dispersion forces, which are the weakest types of intermolecular forces.

02:44 So therefore, since f2 has the weakest intermolecular forces, it has the lowest freezing point out of these three.

02:52 Part c, we want to know which one of these has the lowest vapor pressure at 25 degrees celsius.

02:59 If we remember what vapor pressure is, if we have a liquid, and that liquid, vaporizes into a gas, then that gas exerts a pressure on the inner walls of that container.

03:13 So if we are at the same temperature for three different molecules, then if these molecules have differences in boiling points, then the one with the highest boiling point will not have as much vapor in the air because this is a constant temperature of 25 degrees celsius.

03:37 So if we have a molecule with the lowest boiling point at that constant temperature, then 25 degrees celsius would be closest to that lowest boiling point, and it would have the most vapor particles in the air, and therefore the greatest vapor pressure...

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