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It is possible to change the temperature and pressure of a vessel containing argon gas so that the gas solidifies. (a) What intermolecular forces exist between argon atoms? (b) Is the solid argon a "covalent network solid"? Why or why not?

Short Answer

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(a) The intermolecular forces between argon atoms are van der Waals forces, also known as London dispersion forces or weak inter-atomic forces, arising from temporary fluctuations in electron density. (b) Solid argon is not a covalent network solid because it does not consist of a continuous network of covalent bonds. Instead, it is held together by weak van der Waals forces, resulting in a relatively low melting temperature compared to covalent network solids.

Step by step solution

01

(a) Intermolecular Forces in Argon

Argon, being a noble gas (having a filled valence electron shell) is chemically inert and does not form molecules. Therefore, the intermolecular forces in argon gas are van der Waals forces, also known as London dispersion forces or weak inter-atomic forces. This type of force arises from temporary fluctuations in electron density around the argon atoms, which create instant dipoles that induce dipoles on neighboring atoms.
02

(b) Determine if Solid Argon is a Covalent Network Solid

A covalent network solid is a type of covalent crystal in which the atoms are joined together by a network of covalent bonds. These types of solids are generally very hard, brittle, and have high melting temperatures due to the strength of the covalent bonds in the network. Examples of covalent network solids include diamond, silica, and graphite. Solid argon, on the other hand, is composed of argon atoms held together by the weak van der Waals forces mentioned earlier. These forces are significantly weaker than the covalent bonds present in covalent network solids, resulting in solid argon having a relatively low melting temperature compared to covalent network solids. As a result, solid argon is not considered a covalent network solid because it does not consist of a continuous network of covalent bonds but instead is held together by weak van der Waals forces.

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Most popular questions from this chapter

Acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\), is widely used as an industrial solvent. (a) Draw the Lewis structure for the acetone molecule, and predict the geometry around each carbon atom. (b) Is the acetone molecule polar or nonpolar? (c) What kinds of intermolecular attractive forces exist between acetone molecules? (d) 1-Propanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\), has a molecular weight that is very similar to that of acetone, yet acetone boils at \(56.5^{\circ} \mathrm{C}\) and 1-propanol boils at \(97.2^{\circ} \mathrm{C}\). Explain the difference.

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) melts at \(-114^{\circ} \mathrm{C}\) and boils at \(78^{\circ} \mathrm{C}\). Its density is \(0.789 \mathrm{~g} / \mathrm{mL}\). The enthalpy of fusion of ethanol is \(5.02 \mathrm{~kJ} / \mathrm{mol}\), and its enthalpy of vaporization is \(38.56 \mathrm{~kJ} / \mathrm{mol}\). The specific heats of solid and liquid ethanol are \(0.97 \mathrm{~J} / \mathrm{g}-\mathrm{K}\) and \(2.3 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), respectively. (a) How much heat is required to convert \(25.0 \mathrm{~g}\) of ethanol at \(25^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C} ?\) (b) How much heat is required to convert \(5.00 \mathrm{~L}\) of ethanol at \(-140^{\circ} \mathrm{C}\) to the vapor phase at \(78^{\circ} \mathrm{C}\) ?

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 \(\left(\mathrm{CH}_{3} \mathrm{I}\right)\) : $$ \begin{array}{lcccl} \hline \begin{array}{l} \text { Vapor Pressure } \\ \text { (torr): } \end{array} & \mathbf{1 0 . 0} & \mathbf{4 0 . 0} & \mathbf{1 0 0 . 0} & \mathbf{4 0 0 . 0} \\ \hline T \text { for } \mathrm{CH}_{2} \mathrm{Cl}_{2}\left({ }^{\circ} \mathrm{C}\right): & -43.3 & -22.3 & -6.3 & 24.1 \\ T \text { for } \mathrm{CH}_{3} \mathrm{I}\left({ }^{\circ} \mathrm{C}\right): & -45.8 & -24.2 & -7.0 & 25.3 \\ \hline \end{array} $$ (a) Which of the two substances is expected to have th greater dipole-dipole forces? Which is expected to hav the greater London dispersion forces? Based on your a swers, explain why it is difficult to predict which con pound would be more volatile. (b) Which compour would you expect to have the higher boiling poin Check your answer in a reference book such as the \(C R\) Handbook of Chemistry and Physics. (c) The order volatility of these two substances changes as the ten perature is increased. What quantity must be differer for the two substances in order for this phenomenon occur? (d) Substantiate your answer for part (c) drawing an appropriate graph.

What type of intermolecular force accounts for the following differences in each case? (a) \(\mathrm{CH}_{3} \mathrm{OH}\) boils at \(65^{\circ} \mathrm{C}, \mathrm{CH}_{3} \mathrm{SH}\) boils at \(6{ }^{\circ} \mathrm{C}\). (b) Xe is liquid at atmospheric pressure and \(120 \mathrm{~K}\), whereas Ar is a gas. (c) \(\mathrm{Kr}\), atomic weight 84 , boils at \(120.9 \mathrm{~K}\), whereas \(\mathrm{Cl}_{2}\), molecular weight about 71 , boils at \(238 \mathrm{~K}\). (d) Acetone boils at \(56{ }^{\circ} \mathrm{C}\), whereas 2 -methylpropane boils at \(-12{ }^{\circ} \mathrm{C}\).

Which member of the following pairs has the stronger intermolecular dispersion forces: (a) \(\mathrm{Br}_{2}\) or \(\mathrm{O}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{SH}\), (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl} ?\)

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