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True or false: (a) For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable. (b) For the noble gases the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, dipole- dipole interactions, when present, are always greater than dispersion forces. (d) All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical.

Short Answer

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Based on the analysis of each statement, the short answers are: (a) True (b) False (c) False (d) True

Step by step solution

01

(a) Analyzing the statement

The given statement is "For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable." Dispersion forces, also known as London forces or van der Waals forces, are weak intermolecular attractive forces arising from the instantaneous dipoles created by the random motion of electrons in atoms or molecules. Polarizability refers to the ability of an atom or molecule to form a temporary dipole under the influence of an external electric field. Higher polarizability leads to stronger temporary dipoles and thus stronger dispersion forces. So, this statement is True.
02

(b) Analyzing the statement

The given statement is "For the noble gases, the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table." Dispersion forces are the only type of intermolecular forces acting between noble gas atoms because they are nonpolar. As you go down the periodic table in the noble gas column, the atomic size and polarizability increase. This leads to an increase in the strength of the dispersion forces, which in turn results in an increase in boiling points. Therefore, the statement is False.
03

(c) Analyzing the statement

The given statement is "In terms of the total attractive forces for a given substance, dipole-dipole interactions, when present, are always greater than dispersion forces." Dipole-dipole interactions are stronger intermolecular forces that occur between polar molecules due to their permanent dipoles. However, it is not always true that dipole-dipole forces are greater than dispersion forces. Dispersion forces are always present, and their strength depends on the size and polarizability of the molecules involved. In some cases, dispersion forces can be stronger than dipole-dipole interactions, especially for large, polarizable molecules. So, this statement is False.
04

(d) Analyzing the statement

The given statement is "All other factors being the same, dispersion forces between linear molecules are greater than those between molecules whose shapes are nearly spherical." Linear molecules often have larger surface areas in contact with neighboring molecules, which leads to an increase in the strength of the dispersion forces between them. In contrast, molecules with nearly spherical shapes have smaller surface areas in contact with neighboring molecules, resulting in weaker dispersion forces. So, this statement is True.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dispersion Forces
Dispersion forces, sometimes referred to as London forces, play a crucial role in determining the physical properties of molecules. These are indeed the weakest form of intermolecular interactions, but they are also universal, present in all molecules whether polar or nonpolar. The strength of dispersion forces increases with the size of the molecule and its ability to be polarizable.

For instance, larger molecules have more electrons which can momentarily form instantaneous dipoles, thus resulting in stronger temporary attractions between neighboring molecules. This is why, as pronounced in the textbook exercise, dispersion forces become stronger when the molecules involved are more polarizable, even if they have similar molecular weights.
Polarizability
Polarizability is a measure of how easily the electron cloud around an atom or molecule can be distorted to form a temporary dipole. The electron cloud's response to external electrical fields or the presence of nearby dipoles leads to the formation of temporary attractions.

Heavier atoms with more electrons and looser holds on these electrons tend to be more polarizable. The polarizability of a substance correlates directly with the strength of the dispersion forces that it can exhibit.
  • The ease with which a molecule's electron distribution is distorted dictates its interaction with nearby molecules and contributes to its physical properties.
Dipole-Dipole Interactions
Dipole-dipole interactions arise between molecules that have permanent dipole moments, which means they have ends with partial positive and negative charges. These permanent dipoles attract opposite charges on neighboring molecules, creating a significant intermolecular force.

Although dipole-dipole interactions are generally stronger than dispersion forces, there are exceptions. In cases of highly polarizable molecules with strong dispersion forces, such as large organic molecules, these can sometimes overshadow dipole-dipole interactions.
Van der Waals Forces
Van der Waals forces is an encompassing term that includes both dispersion forces and dipole-dipole interactions. These forces are electric in nature but are different from ionic or covalent bonds in that they are weaker and arise from temporary rather than permanent charge separations.

Understanding van der Waals forces explains many physical phenomena, such as why nonpolar substances can exist in liquid form or why geckos can walk on walls. These forces are influential in nature and technology, affecting how molecules come together and interact.
Boiling Points
The boiling point of a substance is a reflection of the strength of its intermolecular forces. The stronger the intermolecular forces, the more energy is required to separate the molecules into a gaseous state, hence the higher the boiling point.

As seen in the textbook problem, noble gases have higher boiling points as you go down the group in the periodic table due to increased polarizability and stronger van der Waals forces. This trend is opposite to the assertion in the homework problem that 'dispersion forces decrease while boiling points increase,' proving that boiling points are indeed tied to the strength of intermolecular attractions.
Periodic Table Trends
The periodic table is not just a tabulation of elements, it also illustrates trends in their physical and chemical properties. When moving down a group on the periodic table, elements generally become more polarizable due to an increase in atomic size and the number of electrons.

The increasing polarizability ties into the strength of the dispersion forces that is a key factor in determining an element's boiling point. Knowledge of these trends can help predict a substance's behavior without direct experimentation and is therefore indispensable in chemistry.

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Most popular questions from this chapter

(a) How does the average kinetic energy of molecules compare with the average energy of attraction between molecules in solids, liquids, and gases? (b) Why does increasing the temperature cause a solid substance to change in succession from a solid to a liquid to a gas? (c) What happens to a gas if you put it under extremely high pressure?

Ethyl chloride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) boils at \(12^{\circ} \mathrm{C}\). When liquid \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) under pressure is sprayed on a room-temperature \(\left(25^{\circ} \mathrm{C}\right)\) surface in air, the surface is cooled considerably. (a) What does this observation tell us about the specific heat of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g)\) as compared with that of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(l) ?\) (b) Assume that the heat lost by the surface is gained by ethyl chloride. What enthalpies must you consider if you were to calculate the final temperature of the surface?

The fluorocarbon compound \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) has a normal boiling point of \(47.6^{\circ} \mathrm{C}\). The specific heats of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(l)\) and \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}(g)\) are \(0.91\) and \(0.67 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. The heat of vaporization for the compound is \(27.49 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to convert \(35.0 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{~F}_{3}\) from a liquid at \(10.00{ }^{\circ} \mathrm{C}\) to a gas at \(105.00{ }^{\circ} \mathrm{C}\).

Explain why any substance's heat of fusion is generally lower than its heat of vaporization.

(a) Do you expect the viscosity of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\), to be larger or smaller than that of 1-propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\) ? (b) Explain. [Section 11.3]

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