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Describe the intermolecular forces that must be overcome to convert each of the following from a liquid or solid to a gas: (a) \(\mathrm{I}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), (c) \(\mathrm{H}_{2} \mathrm{Se}\).

Short Answer

Expert verified
To convert each compound from a liquid or solid to a gas, the following intermolecular forces must be overcome: (a) For I₂, dispersion forces are present due to its nonpolar nature. (b) For CH₃CH₂OH (ethanol), both dipole-dipole forces and hydrogen bonding are present due to the polar O-H bond. (c) For H₂Se, the primary intermolecular force is dipole-dipole interaction, as hydrogen bonding is not significant due to selenium's lower electronegativity compared to oxygen, nitrogen, or fluorine.

Step by step solution

01

(a) Identify intermolecular forces in I₂

I₂ is a diatomic nonpolar molecule, meaning it has no dipole moment. Therefore, the only type of intermolecular force present in I₂ molecules is dispersion forces (also known as London dispersion forces). These forces are due to instantaneous fluctuations in the electron cloud of a molecule which create temporary dipoles. To convert I₂ from a liquid or solid to a gas, the dispersion forces between the I₂ molecules must be overcome.
02

(b) Identify intermolecular forces in CH₃CH₂OH

CH₃CH₂OH, or ethanol, is a polar molecule. This is due to the presence of the highly polarized oxygen-hydrogen bond (O-H). Therefore, dipole-dipole forces are present between ethanol molecules. Additionally, the O-H bond in ethanol can form hydrogen bonds with other ethanol molecules because oxygen is a highly electronegative element linked to hydrogen. Thus, to convert CH₃CH₂OH from a liquid or solid to a gas, both dipole-dipole forces and hydrogen bonding between ethanol molecules must be overcome.
03

(c) Identify intermolecular forces in H₂Se

H₂Se, or hydrogen selenide, is a polar molecule having a significant dipole moment, due to the difference in electronegativity between hydrogen and selenium atoms. As a result, there are dipole-dipole forces between H₂Se molecules. However, selenium is not as electronegative as oxygen, nitrogen, or fluorine, so hydrogen bonding is not significant in H₂Se. To convert H₂Se from a liquid or solid to a gas, the primary intermolecular force to overcome is the dipole-dipole interaction between H₂Se molecules.

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Key Concepts

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

London Dispersion Forces
London dispersion forces, also known as Van der Waals forces, are the weakest type of intermolecular force. They are present in all molecules, whether polar or nonpolar. These forces arise due to momentary fluctuations in the electron cloud of atoms and molecules, resulting in temporary dipoles. Although each dispersion force is weak, they can collectively have a significant impact on the physical properties of substances.

For example:
  • Even nonpolar molecules like \( \mathrm{I}_2 \) (iodine) exhibit London dispersion forces.
  • These forces are pivotal in understanding why \( \mathrm{I}_2 \) molecules can exist as solids at room temperature before subliming to gases upon heating.
London dispersion forces are essential for stabilizing molecules in solid and liquid phases, though they need to be overcome to transition a substance to the gaseous phase.
Hydrogen Bonding
Hydrogen bonding is a special type of dipole-dipole interaction, but it is much stronger. This bond occurs when a hydrogen atom is covalently bonded to a highly electronegative atom, such as oxygen, nitrogen, or fluorine. These electronegative atoms attract the hydrogen's electron, leaving the hydrogen nucleus exposed and creating a strong attraction to another electronegative atom nearby in a different molecule.

In the case of ethanol (\( \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \)), hydrogen bonds form due to the O-H bond:
  • Oxygen atoms in one ethanol molecule attract the hydrogen atoms of neighboring ethanol molecules.
  • This phenomenon is why ethanol has a higher boiling point compared to non-hydrogen-bonding compounds of similar molecular weight.
  • To transform ethanol from liquid to gas, these robust hydrogen bonds must be broken.
Hydrogen bonding significantly influences the physical properties and boiling points of substances that exhibit this type of intermolecular force.
Dipole-Dipole Interactions
Dipole-dipole interactions occur between polar molecules. These interactions are due to the inherent charge difference in molecules where positive and negative charges attract each other. The strength of dipole-dipole forces depends on the magnitude of the dipole moment, which itself depends on the difference in electronegativity between bonded atoms.

In molecules like hydrogen selenide (\( \mathrm{H}_2\mathrm{Se} \)), dipole-dipole interactions are present:
  • Hydrogen selenide molecules have a polar covalent bond between the slightly positive hydrogen atoms and the more electronegative selenium atom.
  • This creates a dipole where positive ends of one molecule are attracted to negative ends of another, forming an attractive force that holds the molecules together.
  • Overcoming these dipole-dipole interactions is crucial for \( \mathrm{H}_2\mathrm{Se} \) to transition from a liquid or solid state into a gas.
Understanding dipole-dipole interactions helps explain many properties of polar compounds, including their solubility and volatility relative to nonpolar compounds.

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Most popular questions from this chapter

The elements xenon and gold both have solid-state structures with face- centered cubic unit cells, yet Xe melts at \(-112^{\circ} \mathrm{C}\) and gold melts at \(1064^{\circ} \mathrm{C}\). Account for these greatly different melting points.

The following data present the temperatures at which certain vapor pressures are achieved for dichloromethane \(\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)\) and methyl iodide \(\left(\mathrm{CH}_{3} \mathrm{I}\right)\) : $$ \begin{array}{lcccc} \hline \begin{array}{l} \text { Vapor Pressure } \\ \text { (torr): } \end{array} & \mathbf{1 0 . 0} & \mathbf{4 0 . 0} & \mathbf{1 0 0 . 0} & \mathbf{4 0 0 . 0} \\ \hline T \text { for } \mathrm{CH}_{2} \mathrm{Cl}_{2}\left({ }^{\circ} \mathrm{C}\right): & -43.3 & -22.3 & -6.3 & 24.1 \\ T \text { for } \mathrm{CH}_{3} \mathrm{I}\left({ }^{\circ} \mathrm{C}\right): & -45.8 & -24.2 & -7.0 & 25.3 \end{array} $$ (a) Which of the two substances is expected to have the greater dipole-dipole forces? Which is expected to have the greater London dispersion forces? Based on your answers, explain why it is difficult to predict which compound would be more volatile. (b) Which compound would you expect to have the higher boiling point? Check your answer in a reference book such as the \(C R C\) Handbook of Chemistry and Physics. (c) The order of volatility of these two substances changes as the temperature is increased. What quantity must be different for the two substances in order for this phenomenon to occur? (d) Substantiate your answer for part (c) by drawing an appropriate graph.

Sketch a generic phase diagram for a substance that has a more dense solid phase than a liquid phase. Label all regions, lines, and points.

(a) Explain why X-rays can be used to measure atomic distances in crystals but visible light cannot. (b) Why can't \(\mathrm{CaCl}_{2}\) have the same crystal structure as \(\mathrm{NaCl} ?\)

Indicate the type of crystal (molecular, metallic, covalent-network, or ionic) each of the following would form upon solidification: (a) \(\mathrm{CaCO}_{3}\), (b) \(\mathrm{Pt}\), (c) \(\mathrm{ZrO}_{2}\) (melting point, \(2677^{\circ} \mathrm{C}\) ), (d) table sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (e) benzene, (f) \(\mathrm{I}_{2}\).

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